Color and Recall Researchers from the University of British Co...

Question

Color and Recall Researchers from the University of British Columbia conducted trials to investigate the effects of color on the accuracy of recall. Subjects were given tasks consisting of words displayed on a computer screen with background colors of red and blue. The subjects studied 36 words for 2 minutes, and then they were asked to recall as many of the words as they could after waiting 20 minutes. Results from scores on the word recall test are given below. Use a 0.05 significance level to test the claim that variation of scores is the same with the red background and blue background.

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Accepted Answer

Hello, everyone, let's take a look at this question together. A researcher wants to know if the variability in test scores differs between students who take in-person exams and those who take online exams. She collects the following data from two independent samples. Use a 0.05 significance level to test the claim that the variances of the two groups are equal. Assume that the samples are from normally distributed populations, and we know that the in-person. Exams have a sample size of 27, a sample mean of 81.3, and a standard deviation of 7.25, whereas the online exams have a sample size of 24, a sample mean of 78.4, and a standard deviation of 6.12. And the first step in testing the claim that the variances of the two groups are equal, is to conduct our requirement check, which includes that the two populations are independent. We assume that the two samples are simple random samples and that the researcher carefully planned the design for the study so that the selection process follows a random procedure and that each of the two populations is assumed to be normally distributed. And now we can state our hypotheses, which includes a null hypothesis that the sample variance 1 is equal to the sample variance 2 and an alternative hypothesis that Sample variance 1 is not equal to sample variance 2. And next we calculate the test statistic F, which the test statistic F is equal to S21 divided by Squad 2, where we know that S21 is the variance of the group with the larger sample standard deviation and Squad sub 2 is the variance of the group with the smaller sample standard deviation. Which from the information provided in the question, we are given the in person as a sample size of 27 and a standard deviation of 7.25, whereas online has a sample size of 24 and a standard deviation of 6.12. And since 7.25 is Greater than 6.12, we know that Squared sub 1 is the variance of the group with the larger sample standard deviation. So our formula is F is equal to 7.25 square divided by 6.12 squared. So F is approximately 1.403. And now we can determine our degrees of freedom. Which our degrees of freedom 1 is equal to N1 minus 1, so it is 27 minus 1, which equals 26, and our degrees of freedom 2 is equal to N2 minus 1, which is 24 minus 1, giving us 23, so our first degree of freedom is 26. And our second degree of freedom is 23. So then we can find the critical value which for a two-tailed test at alpha equals 0.05, we look up the upper critical value for the calculated degrees of freedom and alpha divided by 2 equals 0.025. So using the table for F distribution with an area of 0.025 in the right tail, since our first degree of freedom, which is 26, is between 24 and 30, we need to perform linear interpolation, which by linear interpolation, we have the formula F sub upper, which is our upper critical value minus 2.2989 divided by 26 minus 24, which equals. 2.2389 minus 2.29 89 divided by 30 minus 24 and rearranging the formula to solve for F upper, we have 2.2389 minus 2.2989 divided by 30 minus 24, multiplied by 26 minus 24 plus 2.2989, which gives us an F upper value of 2.279. Therefore, the upper critical value is 2.279. And since we made a condition that the larger variance is placed in the numerator of the F test statistic, we only need to find the right tailed or upper critical value. And now that we have our upper critical value, we can make our decision, which Is that if our F statistic is greater than 2.279, we reject the null hypothesis. And since 1.403 is less than 2.279, we fail to reject the null hypothesis. So in conclusion, at the 0.05 significance level, there is insufficient evidence to warrant rejection of the claim that the variances of test scores are equal for both groups, which makes answer choice C the correct answer, and all other answer choices are incorrect. I hope you found this video to be helpful. Thank you and goodbye.

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Key Concepts

Hypothesis Testing

Variance and Standard Deviation

ANOVA (Analysis of Variance)

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