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OlgaM077 [116]
3 years ago
10

The mean caloric intake of an adult male is 2800 with a standard deviation of 115. To verify this information, a sample of 25 me

n are selected and determined to have a mean caloric intake of 2950. Determine the 98% confidence interval for mean caloric intake of an adult male. Solution: Since n < 30, a t-test must be used. 2950 – qt(1.98/2,24)*115/sqrt(25) 2950 + qt(1.98/2,24)*115/sqrt(25) [2892.68, 3007.32] What is wrong with this solution?
Mathematics
1 answer:
erma4kov [3.2K]3 years ago
8 0

Answer:

Step-by-step explanation:

We would use the t- distribution.

From the information given,

Mean, μ = 2950

Standard deviation, σ = 115

number of sample, n = 25

Degree of freedom, (df) = 25 - 1 = 24

Alpha level,α = (1 - confidence level)/2

α = (1 - 0.98)/2 = 0.01

We will look at the t distribution table for values corresponding to (df) = 24 and α = 0.01

The corresponding z score is 2.492

We will apply the formula

Confidence interval

= mean ± z ×standard deviation/√n

It becomes

2950 ± 2.492 × 115/√25

= 2950 ± 2.492 × 23

= 2950 ± 57.316

The lower end of the confidence interval is 2950 - 57.316 =2892.68

The upper end of the confidence interval is 2950 + 57.316 = 3007.32

The solution is correct.

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Costs are rising for all kinds of medical care. The mean monthly rent at assisted-living facilities was reported to have increas
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Answer:

a) The 90% confidence interval estimate of the population mean monthly rent is ($3387.63, $3584.37).

b) The 95% confidence interval estimate of the population mean monthly rent is ($3368.5, $3603.5).

c) The 99% confidence interval estimate of the population mean monthly rent is ($3330.66, $3641.34).

d) The confidence level is how sure we are that the interval contains the mean. So, the higher the confidence level, more sure we are that the interval contains the mean. So, as the confidence level is increased, the width of the interval increases, which is reasonable.

Step-by-step explanation:

a) Develop a 90% confidence interval estimate of the population mean monthly rent.

Our sample size is 120.

The first step to solve this problem is finding our degrees of freedom, that is, the sample size subtracted by 1. So

df = 120-1 = 119

Then, we need to subtract one by the confidence level \alpha and divide by 2. So:

\frac{1-0.90}{2} = \frac{0.10}{2} = 0.05

Now, we need our answers from both steps above to find a value T in the t-distribution table. So, with 119 and 0.05 in the t-distribution table, we have T = 1.6578.

Now, we find the standard deviation of the sample. This is the division of the standard deviation by the square root of the sample size. So

s = \frac{650}{\sqrt{120}} = 59.34

Now, we multiply T and s

M = T*s = 59.34*1.6578 = 98.37

The lower end of the interval is the mean subtracted by M. So it is 3486 - 98.37 = $3387.63.

The upper end of the interval is the mean added to M. So it is 3486 + 98.37 = $3584.37.

The 90% confidence interval estimate of the population mean monthly rent is ($3387.63, $3584.37).

b) Develop a 95% confidence interval estimate of the population mean monthly rent.

Now we have that \alpha = 0.95

So

\frac{1-0.95}{2} = \frac{0.05}{2} = 0.025

With 119 and 0.025 in the t-distribution table, we have T = 1.9801.

M = T*s = 59.34*1.9801 = 117.50

The lower end of the interval is the mean subtracted by M. So it is 3486 - 117.50 = $3368.5.

The upper end of the interval is the mean added to M. So it is 3486 + 117.50 = $3603.5.

The 95% confidence interval estimate of the population mean monthly rent is ($3368.5, $3603.5).

c) Develop a 99% confidence interval estimate of the population mean monthly rent.

Now we have that \alpha = 0.99

So

\frac{1-0.95}{2} = \frac{0.05}{2} = 0.005

With 119 and 0.025 in the t-distribution table, we have T = 2.6178.

M = T*s = 59.34*2.6178 = 155.34

The lower end of the interval is the mean subtracted by M. So it is 3486 - 155.34 = $3330.66.

The upper end of the interval is the mean added to M. So it is 3486 + 155.34 = $3641.34.

The 99% confidence interval estimate of the population mean monthly rent is ($3330.66, $3641.34).

d) What happens to the width of the confidence interval as the confidence level is increased? Does this seem reasonable? Explain.

The confidence level is how sure we are that the interval contains the mean. So, the higher the confidence level, more sure we are that the interval contains the mean. So, as the confidence level is increased, the width of the interval increases, which is reasonable.

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