Problem 410  

 

A travel association claims that the mean cost for two adults traveling together on vacation in 115. A random sample of 25 such groups adults has a mean daily meal cast of 120 and a standard deviation of 7.00. Is there enough evidence to reject the claim at α = 0.05?

Complete parts (a) through (d) below.

 

Solution

 

(a)    Write the claim mathematically and identify H_{0} and H_{a}. A null hypothesis H_{0} is a statistical hypothesis that contain a statement of equality, such as ≤ , = , or ≥. The alternative hypothesis H_{a} is the complement of the null hypothesis. It is statement must be true if H_{0} is false and it contain a statement of strict inequality, such as > , ≠ , or <.

 

The appropriate hypothesis are stated below.

H_{0} : ≤ 115

H_{a} : > 115

 

(b)   Calculate the standardized test statistic.

 

First determine which type of test is being conduct. Since the travel association’s claim is that the mean daily meal cost for two adults traveling together on vacation is equal to the suggested amount, this is a two-tailed test.

 

Calculate the standardized test statistic. The standardized test statistic t is given by the formula below, where \bar{x} is the sample mean, µ is the hypothesis mean s is the sample standard deviation, and n is the sample size.

t = \frac{\bar{x}-\mu }{\frac{s }{\sqrt{n}}}

Determine the standardized test statistic, rounding to two decimal places.

t = \frac{\bar{x}-\mu }{\frac{s }{\sqrt{n}}}

= \frac{120 - 115 }{\frac{7 }{\sqrt{25}}}

\approx 3.57

A P-value of a hypothesis test is the probability of obtaining a sample statistic with a value as extreme or more extreme than the one determined from the sample data if the null hypothesis is true. While either the t-distribution table or technology could be used to determine the P-value, for this example, use technology.

Use technology to calculate the P-calculate the P-value for t  \approx 3.57, rounding to three decimal places.

P-value \approx 0.002

 

(c)    If the P-value is less than the level of significance, α = 0.05, reject the null hypothesis.

Otherwise, fail to reject the null hypothesis.

 

Use the information above to interpret the decision in the context if the original claim.

(d)   If the null hypothesis is rejected, than there is sufficient evidence to reject the travel association’s claim that the mean daily coat foe two adults traveling together on vacation is equal to suggested amount. If the null hypothesis is not rejected, than there is not sufficient evidence to reject the claim that the mean daily meal cost for two adults traveling together on vacation is equal to the suggested amount.

 

Use the information above to interpret the decision in the context of the original claim.

 

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