Question

A manufacturer of bedding wants to do quality control on its 400-thread-count sheets. The thread count varies normally with a mean thread count μ = 400 and a standard deviation σ = 8. A quality control researcher randomly selects 16 sheets and finds that the mean thread count for the sample is 395.2. We want to determine if these data provide enough evidence to conclude that the mean thread count is significantly different from the target level. One-sample z-test of μ = 400 vs. μ ≠ 400 Assumed standard deviation = 8 n = 16 Mean = 395.2 z-value = −2.4 Two-tailed p-value = 0.0164 95% confidence interval: (396.0801, 403.9199) Which of the following represents the correct conclusion we can make on the basis of the output (and at the usual significance level of 0.05)?

Answer 1

Answer:

$z=\frac{395.2-400}{\frac{8}{\sqrt{16}}}=-2.4$  

$p_v =2*P(Z<-2.4)=0.0164$  

If we compare the p value and the significance level given $\alpha=0.05$ we see that $p_v<\alpha$ so we can conclude that we have enough evidence to reject the null hypothesis, so we can conclude that the true mean is significantly different from 400.  

$395.2-1.96\frac{8}{\sqrt{16}}=391.28$    

$395.2+1.96\frac{8}{\sqrt{16}}=399.12$

So on this case the 95% confidence interval would be given by (391.28;399.12)

Since the confidence interval not contains the value of 400 we can conclude that the true mean is different from 400 at 5% of significance.      

Step-by-step explanation:

1) Data given and notation  

$\bar X=395.2$ represent the sample mean  

$\sigma=8$ represent the population standard deviation  

$n=16$ sample size  

$\mu_o =7.3$ represent the value that we want to test  

$\alpha=0.05$ represent the significance level for the hypothesis test.  

z would represent the statistic (variable of interest)  

$p_v$ represent the p value for the test (variable of interest)  

2) State the null and alternative hypotheses.  

We need to conduct a hypothesis in order to check if the mean pressure is different from 400, the system of hypothesis are :  

Null hypothesis:$\mu = 400$  

Alternative hypothesis:$\mu \neq 400$  

Since we know the population deviation, is better apply a z test to compare the actual mean to the reference value, and the statistic is given by:  

$z=\frac{\bar X-\mu_o}{\frac{\sigma}{\sqrt{n}}}$ (1)  

z-test: "Is used to compare group means. Is one of the most common tests and is used to determine if the mean is (higher, less or not equal) to an specified value".  

3) Calculate the statistic  

We can replace in formula (1) the info given like this:  

$z=\frac{395.2-400}{\frac{8}{\sqrt{16}}}=-2.4$  

4) P-value  

Since is a two sided test the p value would given by:  

$p_v =2*P(Z<-2.4)=0.0164$  

5) Conclusion  

If we compare the p value and the significance level given $\alpha=0.05$ we see that $p_v<\alpha$ so we can conclude that we have enough evidence to reject the null hypothesis, so we can conclude that the true mean is significantly different from 400.  

6) Confidence interval

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

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

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 tabel to find the critical value. The excel command would be: "=-NORM.INV(0.025,0,1)".And we see that $z_{\alpha/2}=1.96$

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

$395.2-1.96\frac{8}{\sqrt{16}}=391.28$    

$395.2+1.96\frac{8}{\sqrt{16}}=399.12$

So on this case the 95% confidence interval would be given by (391.28;399.12)

Since the confidence interval not contains the value of 400 we can conclude that the true mean is different from 400 at 5% of significance.