"Suppose we are testing the null hypothesis H0: = 20 and the alternative Ha: 20, for a Normal population with = 5. A random sample of 25 observations is drawn from the population, and we find that the sample mean of these observations is = 17.6. The P-value is closest to:"

Complete question is;

Suppose we are testing the null hypothesis H0: μ = 20 and the alternative Ha: μ ≠ 20, for a normal population with σ = 5. A random sample of 25 observations are drawn from the population, and we find the sample mean of these observations is x¯ = 17.6. The P-value is closest to:

a. 0.0668.

b. 0.0082.

c. 0.0164.

d. 0.1336

Option D

Step-by-step explanation:

We are given;

Null hypothesis; H0: μ = 20

Alternative hypothesis; Ha: μ ≠ 20

Population Standard deviation; σ = 5

Sample size; n = 25

Sample mean; x¯ = 17.6

Let's find the z-score from the formula;

z = (x¯ - μ)/(σ/√n)

z = (17.6 - 20)/(5/√5)

z = - 1.073

From online p-value from z-score calculator attached, using z = -1.073; two tailed hypothesis; significance value of 0.05,we have;

The P-Value is 0.283271.

Looking at the given options, the closest to the p-value is option D

what is the original number? you forgot to add it.

total amount of money that he should pay if he has not used his duty free exemption is $98

step-by-step explanation:

given a united states citizen returning to the states declares the following items at the customs office: 3 shirts at $8.50 each 2 dresses at $27.50 each 1 pair of gold cuff links at $17.50 per pair. we have to find how much he pay

cost of 3 shirts if one cost at $8.50 is

cost of 2 dresses if one cost at $27.50 is  

cost of 1 pair of gold cuff at $17.50

so, total amount of money that he should pay if he has not used his duty free exemption is   $25.5+$55+$17.50=$98