Question

You are interested in finding a 90% confidence interval for the average number of names that people can correctly recall after being introduced to 40 people at a party. Based on past research, you know that the standard deviation for this number is 6.34 names. Suppose you surveyed 55 randomly selected people and found that they correctly remembered an average of 18.16 names. The standard deviation for this group was 4.28 names. Round your answers to two decimal places.
A. The sampling distribution follows a __________ distribution.

B. With 90% confidence the mean number of names that all people can remember when introduced to 40 people at a party is between ______ and _______

C. If many groups of 55 randomly selected people were surveyed, then a different confidence interval would be produced for each group. About _____ percent of these confidence intervals will contain the true population mean number of names remembered and about ______ percent will not contain the true population mean number of names remembered.

Answer 1

Answer:

A. The sampling distribution follows a normal distribution

B. With 90% confidence the mean number of names that all people can remember when introduced to 40 people at a party is between 17.04 and 19.28

C. If many groups of 55 randomly selected people were surveyed, then a different confidence interval would be produced for each group. About 90 percent of these confidence intervals will contain the true population mean number of names remembered and about 10 percent will not contain the true population mean number of names remembered  

Step-by-step explanation:

The given parameters are;

The known population standard deviation of the number of names that people can recall after being introduced to 40 people in a party, σ = 6.34 names

The number of people randomly selected in the survey, n = 55 people

The average number of names correctly remembered, $\overline x_1$ = 18.16

The standard deviation of the group surveyed, s = 4.28

A. Based on the Central Limit Theorem CLT, the sampling distribution follows a normal distribution

B. The 90% confidence interval is given as follows;

$C.I.=\bar{x}\pm z_{\alpha /2}\cdot \dfrac{\sigma}{\sqrt{n}}$

Where $z_{\alpha /2}$ at 90% = 1.65, we have;

$C.I.=18.16\pm 1.65 \times \dfrac{4.28}{\sqrt{55}}$

C.I. ≈ 17.20 < μ < 19.11

With 90% confidence the mean number of names that all people can remember when introduced to 40 people at a party is between 17.04 and 19.28

C. The 90% confidence interval gives the certainty that if different groups of 55 randomly selected people were surveyed, then about 90 percent of the confidence intervals will contain the true population mean number of names remembered and about 10 percent will not contain the true population mean number of names remembered.