Question

You are given the sample mean and the population standard deviation. Use this information to construct the​ 90% and​ 95% confidence intervals for the population mean. Interpret the results and compare the widths of the confidence intervals. If​ convenient, use technology to construct the confidence intervals.

A random sample of 60 home theater systems has a mean price of​$131.00. Assume the population standard deviation is​$18.80.

Construct a​ 90% confidence interval for the population mean.
The​ 90% confidence interval is (_____,_____0
​(Round to two decimal places as​ needed.)

Construct a​ 95% confidence interval for the population mean.

The​ 95% confidence interval is(_____,____)
​(Round to two decimal places as​ needed.)


Interpret the results. Choose the correct answer below.
A.
With​ 90% confidence, it can be said that the population mean price lies in the first interval. With​ 95% confidence, it can be said that the population mean price lies in the second interval. The​95% confidence interval is narrower than the​ 90%.
B.
With​ 90% confidence, it can be said that the population mean price lies in the first interval. With​ 95% confidence, it can be said that the population mean price lies in the second interval. The​95% confidence interval is wider than the​ 90%.
C.
With​ 90% confidence, it can be said that the sample mean price lies in the first interval. With​ 95% confidence, it can be said that the sample mean price lies in the second interval. The​ 95% confidence interval is wider than the​ 90%

Answer 1

With​ 90% confidence, it can be said that the population mean price lies in the first interval. With​ 95% confidence, it can be said that the population mean price lies in the second interval. The​ 95% confidence interval is wider than the​ 90%.

Explanation:

We are given that a random sample of 60 home theater systems has a mean price of​$131.00. Assume the population standard deviation is​$18.80.

• Firstly, the pivotal quantity for 90% confidence interval for the population mean is given by;

P.Q. =
`(\bar X-\mu)/((\sigma)/(√(n) ) )` ~ N(0,1)

where,
`\bar X` = sample mean price = $131

`\sigma` = population standard deviation = $18.80

n = sample of home theater = 60

`\mu` = population mean

Here for constructing 90% confidence interval we have used One-sample z test statistics as we know about the population standard deviation.

So, 90% confidence interval for the population mean,
`\mu` is ;

P(-1.645 < N(0,1) < 1.645) = 0.90 {As the critical value of z at 5% level

of significance are -1.645 & 1.645}

P(-1.645 <
`(\bar X-\mu)/((\sigma)/(√(n) ) )` < 1.645) = 0.90

P(
`-1.645 * {(\sigma)/(√(n) ) }` <
`{\bar X-\mu}` <
`1.645 * {(\sigma)/(√(n) ) }` ) = 0.90

P(
`\bar X-1.645 * {(\sigma)/(√(n) ) }` <
`\mu` <
`\bar X+1.645 * {(\sigma)/(√(n) ) }` ) = 0.90

90% confidence interval for
`\mu` = [
`\bar X-1.645 * {(\sigma)/(√(n) ) }` ,
`\bar X+1.645 * {(\sigma)/(√(n) ) }` ]

= [
`131-1.645 * {(18.8)/(√(60) ) }` ,
`131+1.645 * {(18.8)/(√(60) ) }` ]

= [127.01 , 134.99]

Therefore, 90% confidence interval for the population mean is [127.01 , 134.99].

• Now, the pivotal quantity for 95% confidence interval for the population mean is given by;

P.Q. =
`(\bar X-\mu)/((\sigma)/(√(n) ) )` ~ N(0,1)

where,
`\bar X` = sample mean price = $131

`\sigma` = population standard deviation = $18.80

n = sample of home theater = 60

`\mu` = population mean

Here for constructing 95% confidence interval we have used One-sample z test statistics as we know about the population standard deviation.

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

P(-1.96 < N(0,1) < 1.96) = 0.95 {As the critical value of z at 2.5% level

of significance are -1.96 & 1.96}

P(-1.96 <
`(\bar X-\mu)/((\sigma)/(√(n) ) )` < 1.96) = 0.95

P(
`-1.96 * {(\sigma)/(√(n) ) }` <
`{\bar X-\mu}` <
`1.96 * {(\sigma)/(√(n) ) }` ) = 0.95

P(
`\bar X-1.96 * {(\sigma)/(√(n) ) }` <
`\mu` <
`\bar X+1.96 * {(\sigma)/(√(n) ) }` ) = 0.95

95% confidence interval for
`\mu` = [
`\bar X-1.96 * {(\sigma)/(√(n) ) }` ,
`\bar X+1.96 * {(\sigma)/(√(n) ) }` ]

= [
`131-1.96 * {(18.8)/(√(60) ) }` ,
`131+1.96 * {(18.8)/(√(60) ) }` ]

= [126.24 , 135.76]

Therefore, 95% confidence interval for the population mean is [126.24 , 135.76].

Now, with​ 90% confidence, it can be said that the population mean price lies in the first interval. With​ 95% confidence, it can be said that the population mean price lies in the second interval. The ​95% confidence interval is wider than the​ 90%.