The mean number of English courses taken in a two-year time period by male and f
ID: 3313527 • Letter: T
Question
The mean number of English courses taken in a two-year time period by male and female college students is believed to be about the same. An experiment is conducted and data are collected from 29 males and 16 females. The males took an average of two English courses with a standard deviation of 0.7. The females took an average of three English courses with a standard deviation of 0.9. Are the means statistically the same? (Use-005) NOTE: If you are using a Student's t-distribution for the problem, including for paired data, you may assume that the underlying population is normally distributed. (In general, you must first prove that assumption, though.) Part (a) "Part (b) Part (c) Part (d) State the distribution to use for the test. (Enter your answer in the form z or tdf where df is the degrees of freedom. Round your answer to two decimal places.) Part (e) What is the test statistic? (If using the z distribution round your answer to two decimal places, and if using the t distribution round your answer to three decimal places.) Part (f What is the p-value? (Round your answer to four decimal places.) Explain what the p-value means for this problem. lf-tis false, then there is a chance equal to the p-value that the difference in the sample mean number of English courses taken by males and females is at least 1 Olf is true, then there is a chance equal to the p-value that the difference in the sample mean number of English courses taken by males and females is at least 1 O If Ho is true, then there is a chance equal to the p-value that the difference in the sample mean number of English courses taken by males and females is at most 1 O lf is false, then there is a chance equal to the malue that the difference in the sample mean number of English courses taken by males and females is at most 1Explanation / Answer
Solution:-
State the hypotheses. The first step is to state the null hypothesis and an alternative hypothesis.
Null hypothesis: 1 - 2 = 0
Alternative hypothesis: 1 - 2 0
Note that these hypotheses constitute a two-tailed test. The null hypothesis will be rejected if the difference between sample means is too big or if it is too small.
Formulate an analysis plan. For this analysis, the significance level is 0.05. Using sample data, we will conduct a two-sample t-test of the null hypothesis.
Analyze sample data. Using sample data, we compute the standard error (SE), degrees of freedom (DF), and the t statistic test statistic (t).
SE = sqrt[(s12/n1) + (s22/n2)]
SE = 0.25985
DF = 43
t = [ (x1 - x2) - d ] / SE
t = - 3.85
where s1 is the standard deviation of sample 1, s2 is the standard deviation of sample 2, n1 is the size of sample 1, n2 is the size of sample 2, x1 is the mean of sample 1, x2 is the mean of sample 2, d is the hypothesized difference between the population means, and SE is the standard error.
Since we have a two-tailed test, the P-value is the probability that a t statistic having 43 degrees of freedom is more extreme than -3.85; that is, less than -3.85 or greater than 3.85.
Thus, the P-value = 0.00039.
Interpret results. Since the P-value (0.00039) is less than the significance level (0.05), we cannot accept the null hypothesis.
Reject H0, the there is a chance equal to the p-value that the difference in the sample mean number of english courses taken by males and females is atleast 1.
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