Suppose that the delivery times of a nationwide pizza chain are normally distributed with an unknown mean and standard deviation. The delivery times of 26 randomly sampled pizza orders are used to estimate the mean delivery time for the population. What t-score should be used to find the 95% confidence interval for the population mean?

The t-score to be used to find the 95% confidence interval for the population mean is;

Step-by-step explanation:

The t-score which is representative of a point with n - k (n, and k represents the sample size and the number of groups in the test respectively) degrees of freedom  from the Student's T-distribution is a standard test statistic obtained by applying a T-test

Therefore, the t-score used for a confidence interval is obtained by by making use of the the following information;

The number of degrees of freedom = n - k

The confidence level in the confidence interval

The parameters given in the question are;

The number of pizza orders in the sample, n = The sample size = 26

The number of samples used in the estimate, k = 1 group of 26 randomly sampled pizza

The level of confidence (the probability) that the interval contains the population mean = 95%

Therefore, the significance level, α = 1 - 95% = 1 - 95/100 = 0.05

∴ The α-level for the upper and lower limit of the interval are α/2 = 0.05/2 = 0.025 long

The degrees of freedom, df = n - k

∴ For the estimation, DF = 26 - 1 = 25

The t-statistic is therefore the t-value at the intersection of the row for DF = 25 and the 0.025 (one-tail) column (0.025 two-tail column) which gives;

40 minutes

step-by-step explanation:

add the time and multiply by 2