Question

# A customer wants to estimate the average delivery time of a pizza from the local pizza...

A customer wants to estimate the average delivery time of a pizza from the local pizza parlor. The customer wants to obtain a sample mean that falls within 3.212 minutes of the true average delivery time with 90% confidence. If the customer knows the standard deviation of the delivery times of all pizzas is 11.959, how many pizzas will need to be ordered?

Question 6 options:

 1) We do not have enough information to answer this question since we were not given the sample mean.
 2) 43
 3) 37
 4) 48
 5) 38

In a consumer research study, several Meijer and Walmart stores were surveyed at random and the average basket price was recorded for each. It was found that the average basket price for 49 Meijer stores was \$52.202 with a standard deviation of \$11.091. Similarly, 51 Walmart stores had an average basket price of \$65.968 with a standard deviation of \$14.839. If a 99% confidence interval for the difference between the true average basket prices of Meijer versus Walmart is calculated, what is the margin of error? You can assume that the standard deviations of the two populations are statistically similar.

Question 7 options:

 1) 6.904
 2) 6.215
 3) 2.62693
 4) 2.62803
 5) 6.769

a) Since the customer wants to obtain a sample mean that falls within E = 3.212 minutes of the true average delivery time with 90% confidence. If the customer knows the standard deviation of the delivery times of all pizzas is s = 11.959.

To calculate the minimum sample that satisfies the condition at 90% confidence level we need to find the Zc score at 90% confidence level which is =NORM.S.INV(0.95) hence results in Zc = 1.645.

5) N = 38 pizzas need to be ordered.

b) In a consumer research study the given details are:

Assuming the population variances as equal the margin of error is calculated as:

Where 'se' is the standard error is calculated as:

where sp is the pooled variance that is calculated when the population variance is assumed to be equal.

and tc is calculated using excel tool for T-distribution where df( Degree of freedom) is used which is calculated as:

df=n1​+n2​−2=49+51−2=98.

Now the formula used in excel is =T.INV.2T(0.01, 98) this results in tc = 2.63

The se is calculated as:

Now the Margin of error at 99% confidence interval is:

1)

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