Refer to the accompanying data set of mean drive-through service times at dinner in seconds at two fast food restaurants. construct a 95% confidence interval estimate of the mean drive-through service time for restaurant x at dinner; then do the same for restaurant y. compare the results.

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MathPhys MathPhys · Answer 1 · 2020-04-25T01:53:41+02:00

Step-by-step explanation:

Assuming the data is as shown, restaurant X has a mean service time of 180.56, with a standard deviation of 62.6.

The standard error is SE = s/√n = 62.6/√50 = 8.85.

At 95% confidence, the critical value is z = 1.960.

Therefore, the confidence interval is:

180.56 ± 1.960 × 8.85

180.56 ± 17.35

(163, 198)

Restaurant Y has a mean service time of 152.96, with a standard deviation of 49.2.

The standard error is SE = s/√n = 49.2/√50 = 6.96.

At 95% confidence, the critical value is z = 1.960.

Therefore, the confidence interval is:

152.96 ± 1.960 × 6.96

152.96 ± 13.64

(139, 167)

tallinn tallinn · Answer 2 · 2022-04-06T10:16:00+02:00

When 95% confidence interval estimate of the mean drive-through service time for restaurants x and y is = (163, 198) and (139, 167).

We are Assuming the data is as shown, restaurant X has a mean service time of 180.56, with a standard deviation of 62.6.

Then, The standard error is SE = s/√n = 62.6/√50 = 8.85.

After that, At 95% confidence, the critical value is z = 1.960.

Thus, When the confidence interval is:

180.56 ± 1.960 × 8.85

180.56 ± 17.35

(163, 198)

Now, The Restaurant of Y has a mean service time of 152.96, with a standard deviation of 49.2.

When The standard error is SE is = s/√n = 49.2/√50 = 6.96.

After that, At 95% confidence, the critical value is z = 1.960.

Thus, When the confidence interval is:

152.96 ± 1.960 × 6.96

152.96 ± 13.64

(139, 167)

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