1. A research neurologist is testing the effects of a drug on response time

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1 MA 43 WORKSHEET 5 Statistics 1. A research neurologist is testing the effects of a drug on response time by injecting 100 rats with a unit dose of the drug, subjecting each to a neurological stimulus, and recording its response time. The neurologist knows that the mean response time for rats not injected with the drug is 1. seconds. Findings for drug-injected rats are as follows: mean response time = 1.05 seconds with a standard deviation of 0.5 seconds. Set up and test two hypotheses at p<.01. H 0 : The difference in response times of treated and untreated rats is small enough to attribute to sampling error and is not significant. The sample did not show a response rate significantly different from the population, so the drug had no effect on response rate; there was no treatment effect. H 1 : The difference in response times of treated and untreated rats is too great to attribute to sampling error and is significant. The sample did show a response rate significantly different from the population--a treatment effect, and so the drug does affect response time in rats. Show your calculations and your conclusion about the hypotheses. = 1.05 ˆ = μ = 1. ˆ = = = = 100 s = 0.5 s = 0.5 = 0.55 μ ˆ = = = 0.50 Two-tail test Critical value of.660 (df = 99); obtained t is in the critical region. Difference is significant. Accept H 1 ; reject H 1 ; the drug did affect response time.

2 . A sociologist is interested in the relationship between marital status and church attendance. He surveyed a random sample of 173 people who report they never attend church and compared their marital status with the general population. Do people who never attend church differ from the rest of the population in their marital status? Show your calculations. H 0 : Differences between expected and observed frequencies are due to sampling error and are not significant. The sample is not significantly different in marital status from the population, and so people who never attend church do not differ from the rest of the population in their marital status. H 1 : Differences between expected and observed frequencies are greater than would be expected due solely to sampling error and are significant. The sample is significantly different in marital status from the population, and so people who never attend church do differ from the rest of the population in their marital status. There is a treatment effect; church attendance is associated with marital status. Marital Status Pop.% f o f e f o -f e (f o -f e ) (f o -f e ) f e ever Married Separated Divorced Widowed Married Obtained Chi-square = Critical value of Chi-square: (df = k - 1 = 5-1 = 4) Obtained value is OT in critical region, so differences are not significant. Accept H 0 ; reject H 1. o, those who never attend church do not differ from the rest of the population in their marital status.

3 3. A nutritionist, concerned with the problem of teen-age obesity, plans an educational intervention at High Plains High School. He chooses a sample of 65 high school students: 11 seniors, 15 juniors, 19 sophomores, and 0 freshmen. The student body at High Plains is % seniors, 3% juniors, 6% sophomores, and 9% freshmen. Is this sample representative of the student body? Show your calculations. H 0 : The differences between the sample and the population are of a size that is to be expected due to sampling error and are not significant. The sample does not differ significantly from the population in class makeup, and so is representative of the population. H 1 : The differences between the sample and the population are too great to be attributed to sampling error and are significant. The sample differs significantly from the population in class makeup, and so is not representative of the population. Class Pop.% f o f e f o -f e (f o -f e ) (f o -f e ) f e Freshmen Sophomore Junior Senior Obtained Chi-square = Critical value of Chi-square: (df = k - 1 = 3) Obtained value is OT in critical region, so differences are not significant. Accept H 0 ; reject H 1. The sample is representative of the population.

4 4. The time taken by a biological cell to divide is 30 minutes. The population standard deviation is 3.5 minutes. A sample of 16 cells placed in red light was observed to divide in an average time of 31.6 minutes. Did exposure to red light change the dividing time of the cells? Show your work. H 0 : The difference between the dividing time of cells exposed to red light and those not so exposed is no greater than would be expected due to the effect of sampling error and is not significant. The sample and population dividing times are not significantly different, and so exposure to red light does not change the dividing time of cells; there is no treatment effect. H 1 : The difference between the dividing time of cells exposed to red light and those not so exposed is too great to attribute to sampling error and is significant. The sample and population dividing times are significantly different, and so there is a treatment effect; red light changes the dividing time of cells. = 31.6 μ = 30 = 3.5 = 16 = = μ = = = = 1.89 Two-tail test. Critical value of.131 (df = 15) Obtained t is not in critical region. Difference is nonsignificant. Accept H 0 ; reject H 1. o, exposure to red light did not change the dividing time of cells.

5 5. The label on a can of pineapple slices states the mean carbohydrate content per serving is over 50 grams. A random sample of 5 servings has a mean carbohydrate content of 5.3 grams with a standard deviation of 4 grams. Is the company correct in its claim? H 0 : The difference between the carbohydrate content of the sample and the label claim is no greater than would be expected due to sampling error and is not significant. The sample did not contain significantly more than 50 grams of carbohydrate, and so the label claim is inaccurate. H 1 : The difference between the carbohydrate content of the sample and the label claim is too great to attribute to sampling error and is significant. The sample did contain significantly more than 50 grams of carbohydrate, and so the label claim is accurate. = 5.3 μ = 50 = 5 s = 4 s = 16 = = 4.08 ˆ = 4.08 ˆ = = 5 μ ˆ = = =.819 Critical value of (df = 4) Obtained t is in critical region. Difference is significant. Accept H 1 ; reject H 0. The company's claim is accurate.

6 6. High schools in Plains City have a mean dropout rate of 50%. The population standard deviation for all schools is 5%. A sample of 15 Plains City high schools is exposed to a special program designed to reduce the rate of dropping out. At the end of the program the dropout rate for these schools is 47.5%. Write two hypotheses to explain the difference between population and sample dropout rates and test them, showing your calculations. Which hypothesis is accepted? H 0 : The difference between dropout rates in the population and in the sample is no greater than would be expected due solely to sampling error and is not significant. The sample does not have a significantly lower dropout rate than the population, so the program did not reduce the dropout rate; there is no treatment effect. H 1 : The difference between dropout rates in the population and in the sample is too great to attribute to sampling error and is significant. The sample does have a significantly lower dropout rate than the population, so the program reduced the dropout rate; there is a treatment effect. = 47.5 = 5 5 μ = 50 = = = = 5 μ = 15 = = Critical value of (df = 14) Obtained t is in critical region. Difference is significant. Accept H 1 ; reject H 0. The program reduced the dropout rate. OTE: You may have chosen to convert the percentages given in the problem into decimals instead of leaving them as whole numbers; this is acceptable. In that case your sample mean is 0.475; the population mean is

7 0.50; the population standard deviation is 0.05; and the standard error of the mean will be Obtained t will be the same. 7. From his experience a farmer knows that the mean yield of corn per acre on his farm is 150 bushels. When a new seed introduced on the market was tried on 16 randomly-picked experimental 1-acre plots, the mean yield was 158 bushels. Suppose the yield per acre can be assumed to be normally distributed with a standard deviation of yield of 0 bushels. Is the new seed significantly better? H 0 : The difference between the mean farm yield and the test plot yield is no greater than would be expected due to sampling error and is not significant. The test plots did not produce a yield significantly better than the farm's mean yield, and so the new seed is not better. H 1 : The difference between the mean farm yield and test plot yield is too great to attribute to sampling error and is significant. The test plots did produce a yield significantly better than the farm's mean yield, and so the new seed is better. = 158 μ = 150 = 0 = 16 = 0 0 = = = μ = = Critical value of (df = 15) Obtained t is not in critical region. Difference is not significant. Accept H 0 ; reject H 1. The new seed is not significantly better.

8 8. In a sample of 100 college freshmen, the mean IQ is 11. If the mean IQ of the general population is 100 with a standard deviation of 15, can you reject the hypothesis that the mean IQ of college freshmen is no higher than that for the general population? Show your calculations. H 0 : The difference between the mean IQ of college freshmen and the mean IQ of the general population is no greater than would be expected due to sampling error and is not significant. The college freshman mean IQ is not significantly higher than that of the population. H 1 : The difference between the mean IQ of college freshmen and the mean IQ of the general population too great to attribute to sampling error and is significant. The college freshman mean IQ is significantly higher than that of the population. = 11 μ = 100 = 15 = 100 = 15 = = 100 μ = = 0.43 = 7.91 Critical value of (df = 1199) Obtained t is in critical region. Difference is significant. Accept H 1 ; reject H 0. Mean college freshman IQ is higher than that of population. Yes, we can reject the hypothesis (H 0 ) that mean IQ of college freshmen is no higher than that for the general population.

9 9. A sample of 10 political science students completed the Index of Political Liberalism, on which higher scores indicate more liberal beliefs. The mean score was 46.9 with a standard deviation of 15. When the instrument was standardized on the population of political science academics, the mean score was 51. Is this sample of political science students significantly less liberal than the population of political scientists? H 0 : The difference between the mean liberalism scores of political science students and of political scientists is no greater than would be expected due to sampling error and is not significant. Political science students are not less liberal than the population of political scientists. H 1 : The difference between the mean liberalism scores of political science students and of political scientists is too great to attribute to sampling error and is significant. Political science students are less liberal than the population of political scientists. = 46.9 μ = 51 = 10 s = 15 s = 5 = 50 = 15.8 ˆ ˆ = 15.8 = = 10 μ ˆ = = = Critical value of (df = 9) Obtained t is not in critical region. Difference is nonsignificant. Accept H 0 ; reject H 1. Political science students are not less liberal.

10 A larger sample ( = 100) of political science students was assessed, giving a mean of 46.9 with a standard deviation of 15 (same numbers as above). Is this sample of political science students significantly less liberal than the population of political scientists? = 46.9 μ = 51 = 100 s = 15 s = 5 = 7.7 = ˆ ˆ = = = = μ ˆ = = Critical value of (df = 99) Obtained t is in critical region. Difference is significant. Accept H 1 ; reject H 0. Political science students are less liberal.