Mathematics, 24.12.2019 03:31, lala9630

Amanufacturer of piston rings for automobile engines frequently tests the width of the rings for quality control. last week, a random sample of 15 rings were measured, and the mean and standard deviation of the sample were used to construct a 95 percent confidence interval for the population mean width of the rings.

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answered: daniel4349

II and III

See explanation above

Step-by-step explanation:

When all other rings remain the same, which of the following conditions would have resulted in a wider interval than the one constructed?

I. A sample size of 20 with 95 percent confidence

II. A sample size of 15 with 99 percent of confidence

III. A sample size of 12 with 95 percent confidence

Previous concepts

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".

The margin of error is the range of values below and above the sample statistic in a confidence interval.

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

$\bar X$ represent the sample mean

$\mu$ population mean (variable of interest)

s represent the sample standard deviation

n represent the sample size

The confidence interval for the mean is given by the following formula:

$\bar X \pm t_{\alpha/2}\frac{s}{\sqrt{n}}$

The margin of error is given by:

$ME= t_{\alpha/2}\frac{s}{\sqrt{n}}$

I. A sample size of 20 with 95 percent confidence

The original sample size was 15, the degrees of freedom for the original interval would be n-1=14 , the value for $\alpha=0.05$ and $\alpha/2=0.025$ and then the critical value is $t_{critc}=2.14$

And then the Margin of error would be:

$ME=2.14 \frac{s}{\sqrt{15}}=0.553 s$

For the new sample size n=20 the degrees of freedom for the new interval interval would be n-1=19 , the value for $\alpha=0.05$ and $\alpha/2=0.025$ and then the critical value is $t_{critc}=2.09$

And then the Margin error would be:

$ME=2.09 \frac{s}{\sqrt{20}}=0.467 s$

So then we have a lower margin of error so then we will have a shorter interval

II. A sample size of 15 with 99 percent of confidence

The original sample size was 15, the degrees of freedom for the original interval would be n-1=14 , the value for $\alpha=0.05$ and $\alpha/2=0.025$ and then the critical value is $t_{critc}=2.14$

And then the Margin of error would be:

$ME=2.14 \frac{s}{\sqrt{15}}=0.553 s$

For the new interval the sample size n=15 the degrees of freedom for the new interval interval would be n-1=14 , the value for $\alpha=0.01$ and $\alpha/2=0.005$ and then the critical value is $t_{critc}=2.98$

And then the Margin error would be:

$ME=2.98 \frac{s}{\sqrt{15}}=0.769 s$

So then we have a greater margin of error so then we will have a wider interval. And this makes sense since with more confidence the interval needs to be wider.

III. A sample size of 12 with 95 percent confidence

The original sample size was 15, the degrees of freedom for the original interval would be n-1=14 , the value for $\alpha=0.05$ and $\alpha/2=0.025$ and then the critical value is $t_{critc}=2.14$

And then the Margin of error would be:

$ME=2.14 \frac{s}{\sqrt{15}}=0.553 s$

For the new interval the sample size n=12 the degrees of freedom for the new interval interval would be n-1=11 , the value for $\alpha=0.05$ and $\alpha/2=0.025$ and then the critical value is $t_{critc}=2.20$