The trade magazine QSR routinely examines fast-food drive-thru...

Question

The trade magazine QSR routinely examines fast-food drive-thru service times. Their recent research indicates that the mean time a car spends in a McDonald’s drive-thru is 167.1 seconds. A McDonald's manager in Salt Lake City feels that she has instituted a drive-thru policy that results in lower drive-thru service times. A random sample of 70 cars results in a mean service time of 163.9 seconds, with a standard deviation of 15.3 seconds. Determine whether the policy is effective in reducing drive-thru service times.

c. Conduct the appropriate test to determine if the policy is effective.

c. Conduct the appropriate test to determine if the policy is effective.

Accepted Answer

Hi, everyone, let's take a look at this practice problem. This problem says a smartphone manufacturer claims that the average battery life of its new model is at least 20 hours. A consumer group tests a random sample of 16 phones and finds a mean battery life of 18.7 hours with a standard deviation of 2.4 hours. At alpha equal to 0.05, is there sufficient evidence to reject the manufacturer's claim? Assume the population is normally distributed. Now, we're asked to test the claim that the average battery life of these new models of phones is at least 20 hours. So we want to see if the average battery life is greater than or equal to 20 hours. So the first thing we want to do is write down our hypotheses. So for a null hypothesis, H knot, We're going to have that our average battery life is going to be greater than or equal to 20 hours. And our alternative hypothesis, HA is going to be that mu is less than 20 hours. We're also given some other information on the problem. We're told that we have a random sample of 16 phones. So that means that our sample size in is going to be equal to 16. We're also given the sample mean. Which is going to be labeled as X bar, and that's going to be equal to 18.7 hours. And we're also given the sample standard deviation S, which is going to be equal to 2.4 hours. And we're given that alpha is equal to 0.05. Now, note that our alternative hypothesis has the less than symbol in it. So that means that our hypothesis test is going to be left tailed. Also, our population standard deviation is unknown. So, the appropriate test in this case is going to be a one sample T test for the mean. So recall your formula for the Test statistic. That's going to be that T is equal to the quantity of X minus u in quantity, divided by the quantity of S divided by the square root of N in quantity. And we actually have values for all those quantities. So we'll substitute in those values. That means that T is going to be equal to the quantity of 18.7 minus 20 in quantity, divided by the quantity of 2.4 divided by the square root of 16 in quantity. And when we evaluate this expression, this is going to be approximately equal to minus 2.167. So next we need to calculate our criticalt value, and for this we'll first need to find the degrees of freedom. So recall that our degrees of freedom, which we'll label as DF, is equal to N minus 1, which in our case is going to be equal to 16 minus 1, which is equal to 15. So we just need to look up our critical value in our T tables. And so we'll look up for the critical value with 15 degrees of freedom and alpha equal to 0.05, and we see that our critical value TC is going to be equal to minus 1.753. And so now we'll compare our calculated T value to our critical value here, and we see that our calculated T-value, T, is actually less than our critical value T C. That's because minus 2.167 is less than -1.753. Now, since our hypothesis test is left tail, that means that our calculated T value falls within the rejection region, since our calculated T value is less than our critical value. Therefore, we're going to reject our null hypothesis, so we're going to reject H dot. And since we reject our no hypothesis, that means that there is sufficient evidence to reject the manufacturer's claim. So the answer to this problem is going to be yes. There is sufficient evidence to reject the manufacturer's claim. So that is the answer that we were looking for. So I hope that this explanation has been useful, and I'll see you in the next video.

Key Concepts

Hypothesis Testing

One-Sample t-Test

Significance Level and p-Value

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