Ben Shaver

2021-12-30

Here are summary statistic for randomly selected weights of newborn girls: $n=211,\text{}\stackrel{\u2015}{x}=33.7hg,\text{}s=6.8hg.$ Construct a confidence intercal estimate of the mean. Use a 90% confidence level. Are these results very different from the confidence interval $32.1hg<\mu <34.3hg$ with only 16 sample values, $\stackrel{\u2015}{x}=33.2hg$ , and $s=2.6hg$ ?

What is the confidence interval for the population mean$\mu$ ?

Are the results between the two confidence intervals very different?

a) Yes, because the confidence interval limits are not similar.

b) No, because each confidence interval contains the mean of the other confidence interval.

c) No, because the confidence interval limits are similar

d) Yes, because one confidence interval does not contain the mean of the other confidence interval.

What is the confidence interval for the population mean

Are the results between the two confidence intervals very different?

a) Yes, because the confidence interval limits are not similar.

b) No, because each confidence interval contains the mean of the other confidence interval.

c) No, because the confidence interval limits are similar

d) Yes, because one confidence interval does not contain the mean of the other confidence interval.

habbocowji

Beginner2021-12-31Added 22 answers

Step 1

Solution:

From the given information,$n=211,\text{}\stackrel{\u2015}{x}=33.7hg$ and $s=6.8hg.$

Step 2

The confidence level is 0.90.

$1-\alpha =0.90$

$\alpha =1-0.90$

$\alpha =0.10$

Then, a 90% confidence interval for the population mean is

$\stackrel{\u2015}{x}\pm {t}_{\frac{\alpha}{2},\text{}n-1}\left(\frac{s}{\sqrt{n}}\right)=33.7\pm {t}_{\frac{0.10}{2},\text{}211-1}\left(\frac{6.8}{\sqrt{211}}\right)$

$=33.7\pm {t}_{0.05,\text{}210}\left(\frac{6.8}{\sqrt{211}}\right)$

$=33.7\pm \left(1.625\right)(\frac{6.8}{\sqrt{211}}[\text{Using the excel function}\text{}=T.INV(0.05,\text{}210)]$

$=33.7\pm 0.7734$

$=(33.7-0.7734,\text{}33.7+0.07734)$

$=(32.9,\text{}34.5)$

Thus, a 90% confidence interval for the population mean is$32.9hg<\mu <34.5hg$

Step 3

From the given information, the confidence interval for the population mean is$32.1hg<\mu <34.3hg$

From the above two confidence intervals it can be observed that both lower and upper limits of the two confidence intervals are different.

Step 4

Reason for the incorrect options:

From the two confidence intervals it can be observed that limits are not same. To determine the similarity of the two confidence intervals it has to the limits of the intervals.

Thus, the options B, C and D are incorrect.

Solution:

From the given information,

Step 2

The confidence level is 0.90.

Then, a 90% confidence interval for the population mean is

Thus, a 90% confidence interval for the population mean is

Step 3

From the given information, the confidence interval for the population mean is

From the above two confidence intervals it can be observed that both lower and upper limits of the two confidence intervals are different.

Step 4

Reason for the incorrect options:

From the two confidence intervals it can be observed that limits are not same. To determine the similarity of the two confidence intervals it has to the limits of the intervals.

Thus, the options B, C and D are incorrect.

Barbara Meeker

Beginner2022-01-01Added 38 answers

Step 1

Sample size

Sample mean

Standard deviation

we use t distibution as population standard deviation is unknown.

degree of freedom

Significance level,

cratical value

90% Confidence interval

The Two confidence intervals are (32.9, 34.5) and (32.1, 34.3)

The mean 33.2 is included in (32.9, 34.5)

The mean 33.7 is included in (32.1, 34.3)

So each confidence interval contains the mean of the other confidence interval.

karton

Expert2022-01-04Added 613 answers

Our t-statistic is less than 1.97. So,

We have to accept Noll Hypothesy

There is no difference brom the confidence interval brom the two Sample

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