A travel analyst claims the mean daily base price for renting ...

Question

A travel analyst claims the mean daily base price for renting a full-size or less expensive vehicle in Vancouver, British Columbia, is more than $86. You want to test this claim. In a random sample of 40 full-size or less expensive vehicles available to rent in Vancouver, British Columbia, the mean daily base price is $93.23. Assume the population standard deviation is $28.90. At α=0.10, do you have enough evidence to support the analyst’s claim?

Accepted Answer

All right. Hello, everyone. So this question says, a food critic claims that the main costs of a dinner meal at casual restaurants in the city is more than $82. You take a random sample of 40 such restaurants and find that the mean meal price is $88.60. Assume that the population standard deviation is $25.20 and the distribution of meal prices is normal. At the 0.10 significance level, is there enough evidence to support the critics claim? All right, so here we're testing a claim about a population mean with a known population standard deviation, which means that a Z test for the mean is appropriate in this case. So Because the question is asking if we can support the claim that the critic is making, the claim, therefore, corresponds to the alternative hypothesis. So each of 1. would state that me, the population mean, is greater than 82. The the null hypothesis on the other hand, would state that mu is equal to 82. Now, because the alternative hypothesis states that a variable is greater than a certain number, this is a right-tailed test. And you can think of it as the variable being greater than or to the right of a certain number in a number line. So now let's identify the information that we have. We know that mu knot, which is the stated mean, is equal to 82. And we also know that the sample mean X bar is equal to 88.60. The population standard deviation sigma. Is equal to 25.20. The sample size N is equal to 40. And alpha equals 0.10. So with this information, we can now calculate the test statistic. And we can use the Zest formula for the population mean, because sigma is known to us. So Z Is equal to x bar subtracted by mu knot. Divided by sigma divided by the square root of N. So, substituting the values that you have, Z is equal to 88.60. Subtracted by 82. Divided by 25.20 divided by the square root of 40. And this equals approximately 1.66. Now you're left to find the critical value which you would then compare to the test statistic. So because this is a right-tailed test. At alpha equals 0.10, the critical Z value is equal to approximately 1.28. And now, notice how our test statistic 1.66 is greater than our critical value. Because this is a right-tailed test, and our critical value, excuse me, our test statistic is greater than our critical value. It falls within the rejection region. Which means that we can reject. The null hypothesis. So what does that tell you now? About the conclusion you can make. Well, rejecting the null hypothesis states there is sufficient evidence at the 0.10 significance level to support the critics claim that the main costs of a dinner meal is more than $82. So going back to the overall question of if there is enough evidence to support the critics claim, your answer. Would be yes. And there you have it. So with that being said, thank you so very much for watching and I hope you found this helpful.

Key Concepts

Hypothesis Testing

P-Value

Confidence Intervals

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