(1 point) The with a mean of 30 seconds and a standard deviation of 2.1 seconds.
ID: 3340424 • Letter: #
Question
(1 point) The with a mean of 30 seconds and a standard deviation of 2.1 seconds. You wish to see if the mean time is changed by vigorous exercise, so you have a group of 20 college students exercise vigorously for 30 minutes and then complete the maze. It takes them an average of z = 29.3 seconds to complete the maze. Use this information to test the hypotheses H0 : = 30 Ha : +30 Conduct a test using a significance level of 0.01. time needed for college students to complete a certain paper-and-pencil maze follows a normal distribution (a) The test statistic (b) The positive critical value, 20 = The final conclusion is OA. There is enough evidence to support the claim. O B. There is not enough evidence to support the claim..Explanation / Answer
Solution:-
State the hypotheses. The first step is to state the null hypothesis and an alternative hypothesis.
Null hypothesis: = 30
Alternative hypothesis: 30
Note that these hypotheses constitute a two-tailed test. The null hypothesis will be rejected if the sample mean is too big or if it is too small.
Formulate an analysis plan. For this analysis, the significance level is 0.01. The test method is a one-sample z-test.
Analyze sample data. Using sample data, we compute the standard error (SE),z statistic test statistic (z).
SE = s / sqrt(n)
S.E = 0.4696
z = (x - ) / SE
z = - 1.49
zcritical = + 2.575
where s is the standard deviation of the sample, x is the sample mean, is the hypothesized population mean, and n is the sample size.
Since we have a two-tailed test, the P-value is the probability that the z statistic is less than -1.49 or greater than 1.49.
Thus, the P-value = 0.1362
Interpret results. Since the P-value (0.1362) is greater than the significance level (0.01), we cannot reject the null hypothesis.
There is not enough evidence to support the claim.
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