Question

A simple random sample of 60 items resulted in a sample mean of 80. The population standard deviation is σ = 15. a. Compute the 95% confidence interval for the population mean (to 1 decimal). ( __________ , __________ ) b. Assume that the same sample mean was obtained from a sample of 120 items. Provide a 95% confidence interval for the population mean (to 2 decimals). ( __________ , __________ ) c. What is the effect of a larger sample size on the margin of error? _________________

Answer 1

Answer:

(a) 95% confidence interval for the population mean is [76.2 , 83.8].

(b) 95% confidence interval for the population mean after change in sample size is [77.32 , 82.68].

(c) The effect of a larger sample size on the margin of error is that the margin of error will increase.

Step-by-step explanation:

We are given a simple random sample of 60 items resulted in a sample mean of 80. The population standard deviation is σ = 15.

(a) Firstly, the pivotal quantity for 95% confidence interval for the  population mean is given by;

                           P.Q. = $\frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }$  ~ N(0,1)

where, $\bar X$ = sample mean = 80

            $\sigma$ = population standard deviation = 15

            n = sample of items = 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 < $\frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }$ < 1.96) = 0.95

P( $-1.96 \times {\frac{\sigma}{\sqrt{n} } }$ < ${\bar X-\mu}$ < $1.96 \times {\frac{\sigma}{\sqrt{n} } }$ ) = 0.95

P( $\bar X-1.96 \times {\frac{\sigma}{\sqrt{n} } }$ < $\mu$ < $\bar X+1.96 \times {\frac{\sigma}{\sqrt{n} } }$ ) = 0.95

95% confidence interval for $\mu$ = [ $\bar X-1.96 \times {\frac{\sigma}{\sqrt{n} } }$ , $\bar X+1.96 \times {\frac{\sigma}{\sqrt{n} } }$ ]

                                                 = [ $80-1.96 \times {\frac{15}{\sqrt{60} } }$ , $80+1.96 \times {\frac{15}{\sqrt{60} } }$ ]

                                                 = [76.2 , 83.8]

Therefore, 95% confidence interval for the population mean is [76.2 , 83.8].

(b) Now, the sample size has been changed to 120 items.

So, 95% confidence interval for $\mu$ = [ $\bar X-1.96 \times {\frac{\sigma}{\sqrt{n} } }$ , $\bar X+1.96 \times {\frac{\sigma}{\sqrt{n} } }$ ]

                                                 = [ $80-1.96 \times {\frac{15}{\sqrt{120} } }$ , $80+1.96 \times {\frac{15}{\sqrt{120} } }$ ]

                                                 = [77.32 , 82.68]

Therefore, 95% confidence interval for the population mean after change in sample size is [77.32 , 82.68].

(c) The effect of a larger sample size on the margin of error is that the margin of error will increase that's why the confidence interval in part (b) is narrower than part (a).

Answer 2

The larger the sample size, the lower the margin of error.

What is confidence interval?

Confidence interval shows the range of values for which a random sample might be.

a) The z score of 95% confidence interval is 1.96. sample = 60, mean = 80 and standard deviation = 15. Hence, margin of error (E) is:

$E=z_\frac{\alpha }{2} *\frac{standard\ deviation}{\sqrt{sample\ size} } =1.96*\frac{15}{\sqrt{60} } =3.8$

The confidence interval = mean ± E = 80 ± 3.8 = (76.2, 83.8)

b) The z score of 95% confidence interval is 1.96. sample = 120, mean = 80 and standard deviation = 15. Hence, margin of error (E) is:

$E=z_\frac{\alpha }{2} *\frac{standard\ deviation}{\sqrt{sample\ size} } =1.96*\frac{15}{\sqrt{120} } =2.68$

The confidence interval = mean ± E = 80 ± 2.68 = (77.32, 82.68)

The larger the sample size, the lower the margin of error.

Find out more on confidence interval at: brainly.com/question/25779324