Question

Suppose the weights of seventh‑graders at a certain school vary according to a Normal distribution, with a mean of 100 pounds and a standard deviation of 7.5 pounds. A researcher believes the average weight has decreased since the implementation of a new breakfast and lunch program at the school. She finds, in a random sample of 35 students, an average weight of 98 pounds. What is the P ‑value for an appropriate hypothesis test of the researcher’s claim?

Answer 1

Answer:

We conclude that there is no change in the average weight since the implementation of a new breakfast and lunch program at the school.

Step-by-step explanation:

We are given the following in the question:

Population mean, μ = 100 pounds

Sample mean, $\bar{x}$ = 98 pounds

Sample size, n = 35

Alpha, α = 0.05

Population standard deviation, σ = 7.5 pounds

First, we design the null and the alternate hypothesis

$H_{0}: \mu = 100\text{ pounds}\\H_A: \mu < 100\text{ pounds}$

We use one-tailed(left) z test to perform this hypothesis.

Formula:

$z_{stat} = \displaystyle\frac{\bar{x} - \mu}{\frac{\sigma}{\sqrt{n}} }$

Putting all the values, we have

$z_{stat} = \displaystyle\frac{98 - 100}{\frac{7.5}{\sqrt{35}} } = -1.577$

Now, $z_{critical} \text{ at 0.05 level of significance } = -1.64$

We calculate the p-value with the help of standard normal table.

P-value = 0.057398

Since the p-value is greater than the significance level, we fail to reject the null hypothesis and accept is.

We conclude that there is no change in the average weight since the implementation of a new breakfast and lunch program at the school.