Xom Phim Onine, Phim Viets x Student/Assignment Extra Quiz t?dep-175431E a3 My N
ID: 3313597 • Letter: X
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Xom Phim Onine, Phim Viets x Student/Assignment Extra Quiz t?dep-175431E a3 My Nobes Ask Your 11/14 peints 1 The mean number of English courses taken in a two-year time period are collected from 29 males and 16 females. The males took an average of two English courses w courses with a standard deviation of 0.9. Are the means statistically the same? (use a . 0.05) 3. Previous Answers by male and female college students is believed to be about the same. An experiment is conducted with a standard deviation of 0.7. The females took an average of three English data, you may assume that the underlying population is normally distributed. (In general, you NOTE: If you are using a Student's t-distribution for the problem, including for paired must first prove that assumption, though.) Part (a) Part (c) Part (d your answer to two decimal places) State the distribution to use for the test. (Enter your answer in the form zortof where disthe degrees or freedom. Rend y 2.02 x Part (e) Part (0 What is the p-value? (Round your answer to four decimal places.) 000039 X Explain what the p-value means for this problem Ho is talse, then there is a chance equal to the p-value that the difference in the sample mean number of English counses taken by males and females is at least 1 is a chance equal to the pvalue that the difference in the sample mean number of English courses taken by males and females is at least t then there is a chance equal to the p-value that the ifference in the sample mean number of English courses takan by males and temales is at most 1 courses taken by males and females most 1 s a chance equal to the p-value that the oifference in the sample mean number of English courses taken by males and females is at most 1,Explanation / Answer
t43 (43 degrees of freedom because males=29-1=28 and females=16-1=15, so 28+15=43)
Sp^2= (n1-1)S1^2+(n2-1)S2^2/(n1-1)+(n2-1)
= (29-1)*0.7^2+(16-1)*0.9^2/28+15
= 0.6016
tSTAT=(X1-X2)-(µ1-µ2)/Sp^2(1/n1+1/n2)
=(2-3)-0/0.6016(1/29+1/16)
=-1/0.2415
=-4.14
The P-Value is .000159.
The result is significant at p < .05.
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