Question

# DATA: According to data, 72.4% of the population in Kahului, Maui drive alone for their commute...

DATA: According to data, 72.4% of the population in Kahului, Maui drive alone for their commute to work. You believe this is too high and will perform a study at a 5% level of significance.

A group of 30 people were sampled in Kahului, Maui where 21 people stated that they drive alone for their commute to work.

H0: p = 0.724
Ha: p < 0.724

pcap = 21/30 = 0.7

Test statistic,
z = (pcap - p)/(p*(1-p)/n))
z = (0.7 - 0.724)/(0.724 - (1-0.724)/30)
z = -0.03

p-value = 0.488
As p-value > 0.05, fail to reject H0

As this is left tailed test, critical value is to the left of standard curve critical value = -1.64

Question: Graph the rejection region, p-value and critical value. (Create the normal distribution curve) Explain why we fail to reject the null hypothesis, the significance of this and what the p value/ critical value indicate.

Solution :

This is the left tailed test .

The null and alternative hypothesis is

H0 : p = 0.724

Ha : p < 0.724

= x / n = 21 / 30 = 0.7

P0 = 0.724

1 - P0 = 0.276

Test statistic = z

= - P0 / [P0 * (1 - P0 ) / n]

= 0.7 - 0.724 / [(0.724 * 0.276) / 30]

= -0.294

P(z < -0.29) = 0.3859

P-value = 0.3859

= 0.05

Critical value = -1.645

P-value >

Fail to reject the null hypothesis .

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