Mathematics, 07.04.2020 18:52 youngchapo813p8d9u1

A social researcher claims that the average adult listens to the radio less than 26 hours per week. He collects data on 25 individuals' radio listening habits and finds that the mean number of hours that the 25 people spent listening to the radio was 22.4 hours. If the population standard deviation is known to be eight hours, can we conclude at the 1% significance level that he is right

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There is not sufficient evidence to support the claim that average adult listens to the radio less than 26 hours per week at 0.01 significance level.

Step-by-step explanation:

We are given the following in the question:

Population mean, μ = 26 hours per week

Sample mean, $\bar{x}$ = 22.4 hours per week

Sample size, n = 25

Alpha, α = 0.01

Population standard deviation, σ = 8 hours per week

First, we design the null and the alternate hypothesis

$H_{0}: \mu = 26\text{ hours per week}\\H_A: \mu < 26\text{ hours per week}$

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

Formula:

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

Putting all the values, we have

$z_{stat} = \displaystyle\frac{22.4 - 26}{\frac{8}{\sqrt{25}} } = -2.25$

Now, we calculate the p-value from the table.

P-value = 0.0122

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

Conclusion:

Thus, there is not sufficient evidence to support the claim that average adult listens to the radio less than 26 hours per week at 0.01 significance level.

Answer from: Quest

The answer would be a

Answer from: Quest

step-by-step explanation:

i found the life tables for the year 2012.

life table of males at age 30. probability of dying between ages x and x+1

total population:

29–30 . . . . . . . . . . . . 0.001030

30–31 . . . . . . . . . . . . 0.001061

males:

29–30 . . . . . . . . . . . . 0.001420

30–31 . . . . . . . . . . . . 0.001448

white population

29–30 . . . . . . . . . . . . 0.001005

30–31 . . . . . . . . . . . . 0.001037

white males

29–30 . . . . . . . . . . . . 0.001372

30–31 . . . . . . . . . . . . 0.001399

non hispanic white population

29–30 . . . . . . . . . . . . 0.001083

30–31 . . . . . . . . . . . . 0.001116

non hispanic white males

29–30 . . . . . . . . . . . . 0.001477

30–31 . . . . . . . . . . . . 0.001503

jacob, non hispanic white male. 30 years old.

probability that jacob lives to be 30 years old.

since the given probabilities are probabilities of dying, i think we simply subtract it from 1 to get the probability of living. we will use the probability of non hispanic white male.

probability of living to be 30 y.o.

29 - 30 : 1 - 0.001477 = 0.998523

30 - 31 : 1 - 0.001503 = 0.998497

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A social researcher claims that the average adult listens to the radio less than 26 hours per week....

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