Mathematics, 22.04.2020 03:39 lwaites18

Sample standard deviation of the breaking strengths of these welds were found to be 94 ksi (thousand pounds per square inch) and 3 ksi, respectively. (a) Construct a 95% confidence interval for the true mean breaking strength of such welds. (b) Construct a 99% confidence interval for the true mean breaking strength of such welds. (c) Is there evidence that the true mean breaking strength of such welds is less than 95 ksi

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Answer from: lizzy2951

a) $94-2.01\frac{3}{\sqrt{50}}=93.15$

$94+2.01\frac{3}{\sqrt{50}}=94.85$

So on this case the 95% confidence interval would be given by (93.15;94.85)

b) $94-2.68\frac{3}{\sqrt{50}}=92.86$

$94+2.68\frac{3}{\sqrt{50}}=95.14$

So on this case the 99% confidence interval would be given by (92.86;95.14)

c) At 95% of confidence we have that the upper limit is lower than 95 so then at this level we have enough evidence to conclude that the true mean is lower than 95 ksi, but for 99% of confidence we have the opposite result.

Step-by-step explanation:

Assuming this complete question: Weld Breaking Strength In an inertia welding experiment, a sample of 50 welds were tested. The sample mean and sample standard deviation of the breaking strengths of these welds were found to be 94 ksi (thousand pounds per square inch) and 3 ksi, respectively.

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 for the sample

$\mu$ population mean (variable of interest)

s represent the sample standard deviation

n represent the sample size

(a) Construct a 95% confidence interval for the true mean breaking strength of such welds.

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

$\bar X \pm t_{\alpha/2}\frac{s}{\sqrt{n}}$ (1)

In order to calculate the critical value $t_{\alpha/2}$ we need to find first the degrees of freedom, given by:

df=n-1=50-1=49

Since the Confidence is 0.95 or 95%, the value of $\alpha=0.05$ and $\alpha/2 =0.025$ , and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-T.INV(0.025,49)".And we see that $t_{\alpha/2}=2.01$

Now we have everything in order to replace into formula (1):

$94-2.01\frac{3}{\sqrt{50}}=93.15$

$94+2.01\frac{3}{\sqrt{50}}=94.85$

So on this case the 95% confidence interval would be given by (93.15;94.85)

(b) Construct a 99% confidence interval for the true mean breaking strength of such welds.

Since the Confidence is 0.99 or 99%, the value of $\alpha=0.01$ and $\alpha/2 =0.005$ , and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-T.INV(0.005,49)".And we see that $t_{\alpha/2}=2.68$

Now we have everything in order to replace into formula (1):

$94-2.68\frac{3}{\sqrt{50}}=92.86$

$94+2.68\frac{3}{\sqrt{50}}=95.14$

So on this case the 99% confidence interval would be given by (92.86;95.14)

Part c

At 95% of confidence we have that the upper limit is lower than 95 so then at this level we have enough evidence to conclude that the true mean is lower than 95 ksi, but for 99% of confidence we have the opposite result.

(c) Is there evidence that the true mean breaking strength of such welds is less than 95 ksi? Use the five- step hypothesis testing procedure in answering this question. The confidence level is 95%. (Optional)

Answer from: Quest

Not enough information

Answer from: Quest

Dbecause there are negatives surrounding ding it
Which expression has a negative value

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Sample standard deviation of the breaking strengths of these welds were found to be 94 ksi (thousand...

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