The time needed for college students to complete a certain paper-and-pencil maze

Question

The time needed for college students to complete a certain paper-and-pencil maze follows a normal distribution with a mean of 30 seconds and a standard deviation of 3.3 seconds. You wish to see if the mean time mu is changed vigorous exercise so you have a group of 15 college students exercise vigorously for 30 minutes and men complete me maze it lakes them an average of x 27.4 seconds to complete the maze use this information to test the hypotheses H_0: mu = 30 Conduct a test using a significance level of a = 0.05 The test statistic The positive critical value z^* = The final conclusion is A. There is not sufficient evidence to reject the null hypothesis that mu = 30 B. We can reject the null hypothesis that mu = 30 and accept that mu notequalto 30.

Explanation / Answer

Solution:

Part a

The test statistic formula is given as below:

Z = (Xbar - µ) / [/sqrt(n)]

We are given

Xbar = 27.4

µ = 30

= 3.3

n = 15

Z = (27.4 – 30) / [ 3.3/sqrt(15)] = -3.0514

The test statistic = Z = -3.0514

Part b

Lower critical value = -1.96

Upper critical value = 1.96

Positive critical value = 1.96

Part c

P-value = 0.0023

= 0.05

P-value <

So, we reject the null hypothesis

We can reject the null hypothesis that µ = 30 and accept that µ 30.

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