Question

1).Calculate T, when sample mean is 120, population mean is 100, standard deviation is 20 and smaple size is 10 using levels of confidence at 0.05. 2). Determine if the sample mean is significantly different from the population mean.

Answer 1

Answer:

$t=\frac{120-100}{\frac{20}{\sqrt{10}}}=3.16$  

$p_v =2*P(t_{9}>3.16)=0.012$  

If we compare the p value and a significance level assumed $\alpha=0.05$ we see that $p_v<\alpha$ so we can conclude that we can reject the null hypothesis, and the true mean is significantly different from 100 at 5% of significance.  

Step-by-step explanation:

Data given and notation

$\bar X=120$ represent the sample mean  

$s=20$ represent the standard deviation for the sample

$n=10$ sample size  

$\mu_o =100$ represent the value that we want to test  

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

t would represent the statistic (variable of interest)  

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

State the null and alternative hypotheses to be tested  

We need to conduct a hypothesis in order to determine if the mean is different from 100, the system of hypothesis would be:  

Null hypothesis:$\mu = 100$  

Alternative hypothesis:$\mu \neq 100$  

Compute the test statistic  

We don't know the population deviation, so for this case is better apply a t test to compare the actual mean to the reference value, and the statistic is given by:  

$t=\frac{\bar X-\mu_o}{\frac{s}{\sqrt{n}}}$ (1)  

t-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".  

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

$t=\frac{120-100}{\frac{20}{\sqrt{10}}}=3.16$  

Now we need to find the degrees of freedom for the t distirbution given by:

$df=n-1=10-1=9$

Conclusion

Since is a tao tailed test the p value would be:  

$p_v =2*P(t_{9}>3.16)=0.012$  

If we compare the p value and a significance level assumed $\alpha=0.05$ we see that $p_v<\alpha$ so we can conclude that we can reject the null hypothesis, and the true mean is significantly different from 100 at 5% of significance.  

Answer 2

Answer:

The sample mean is significantly different from the population mean.

Step-by-step explanation:

Null hypothesis: The sample mean is the same as the population mean.

Alternate hypothesis: The sample is significantly different from the population mean

Test statistic (t) = (sample mean - population mean) ÷ (sd/√n) = (120 - 100) ÷ (20/√10) = 20 ÷ 6.32 = 3.16

n = 10

degree of freedom = n - 1 = 10 - 1 = 9

From the t-distribution table, critical value corresponding to 9 degrees of freedom and 5% significance level is 2.262

The test is a two tailed, therefore the region of no rejection of the null hypothesis lies between -2.262 and 2.262.

Conclusion:

Reject the null hypothesis because the test statistic 3.16 lies outside the region bounded by the critical values.

The sample mean is significantly different from the population mean.