Mathematics, 27.05.2020 21:04 JvGaming2001

A survey of the age of people denied promotion was conducted. In a random sample of 23 people the average age was 47.0 with a sample standard deviation of 7.2 Assume this sample comes from a population that is normally distributed. a. Calculate the EBM then construct a 95% confidence interval for the average of the age of people denied (round the values of the interval to 1 decimal place). b. If a manager claimed that the average age of people denied promotion was 51, would you agree or disagree based on your confidence interval? Explain your reasoning.

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Answer from: kaylienguyen

(a) 95% confidence interval for the average of the age of people denied is [43.9 , 50.1].

(b) Based on our confidence interval, we would disagree with the manager's claim.

Step-by-step explanation:

We are given that a survey of the age of people denied promotion was conducted. In a random sample of 23 people the average age was 47.0 with a sample standard deviation of 7.2.

Assume this sample comes from a population that is normally distributed.

Firstly, the Pivotal quantity for 95% confidence interval for the population mean is given by;

P.Q. = $\frac{\bar X-\mu}{\frac{s}{\sqrt{n} } }$ ~ t_n_-_1

where, $\bar X$ = sample average age = 47.0

$\sigma$ = sample standard deviation = 7.2

n = sample of people = 23

$\mu$ = true average of the age of people denied

Here for constructing 95% confidence interval we have used One-sample t test statistics as we don't know about population standard deviation.

So, 95% confidence interval for the population mean, $\mu$ is ;

P(-2.074 < t_1_0_0 < 2.074) = 0.95 {As the critical value of t at 22 degree

of freedom are -2.074 & 2.074 with P = 2.5%}

P(-2.074 < $\frac{\bar X-\mu}{\frac{s}{\sqrt{n} } }$ < 2.074) = 0.95

P( $-2.074 \times {\frac{s}{\sqrt{n} } }$ < ${\bar X-\mu}$ < $2.074 \times {\frac{s}{\sqrt{n} } }$ ) = 0.95

P( $\bar X-2.074 \times {\frac{s}{\sqrt{n} } }$ < $\mu$ < $\bar X+2.074 \times {\frac{s}{\sqrt{n} } }$ ) = 0.95

(a) 95% confidence interval for $\mu$ = [ $\bar X-2.074 \times {\frac{s}{\sqrt{n} } }$ , $\bar X+2.074 \times {\frac{s}{\sqrt{n} } }$ ]

= [ $47-2.074 \times {\frac{7.2}{\sqrt{23} } }$ , $47+2.074 \times {\frac{7.2}{\sqrt{23} } }$ ]

= [43.9 , 50.1]

Therefore, 95% confidence interval for the average of the age of people denied is [43.9 , 50.1].

(b) Based on our confidence interval, we would disagree with the manager's claim that the average age of people denied promotion was 51 because the interval doesn't contain the value 51.

Answer from: Quest

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Answer from: Quest

step-by-step explanation:

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