You are asking a question I had in 6th grade ma dood
The answer to your question is eligibility for the national merit scholarship program
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)
