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Gnoma [55]
4 years ago
3

A recent national survey found that parents read an average (mean) of 10 books per month to their children under five years old.

The population standard deviation is 5. The distribution of books read per month follows the normal distribution. A random sample of 25 households revealed that the mean number of books read last month was 12. At the .01 significance level, can we conclude that parents read more than the average number of books to their children? What is the probability of a Type II error?
Mathematics
1 answer:
Masja [62]4 years ago
6 0

Answer:

Decision: Not reject the null hypothesis.

β = 0.0505

Step-by-step explanation:

Hello!

The study variable is X: "books read per month to kids under 5 years old by their parents."

This variable follows a normal distribution

X~N(μ;σ²)

and the population standard deviation is known σ= 5

The objective of this exercise is to test if the average of books read per month is greater than 10 books per month. Symbolically μ > 10 Remember: the null hypothesis always carries the = symbol, with this in mind the hypothesis are:

H₀: μ = 10

H₁: μ > 10

α: 0.01

Since the study variable has a normal distribution and the population variance is known, the best statistic to use is:

Z= <u> X[bar] -  μ </u>~N(0;1)

        σ/√n

The critical region is one-tailed to the right (positive)

Z_{1-\alpha } = Z_{0.99} = 2.33

If Z ≥ 2.33, you reject the null hypothesis.

If Z < 2.33, you do not reject the null hypothesis.

Z= <u> X[bar] -  μ </u>

        σ/√n

Z= <u> 12 -  10  </u> = 2

     5/√25

The decision is to not reject the null hypothesis, this means, that the average number of books parents read to their children per month is 10 or less.

The probability of a Type II error is β = 0.0505

I hope this helps!

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