Question

A hypothesis regarding the weight of newborn infants at a community hospital is that the mean is 6.6 pounds. A sample of seven infants is randomly selected and their weights at birth are recorded as 9.0, 7.3, 6.0, 8.8, 6.8, 8.4, and 6.6 pounds. What is the decision for a statistical significant change in average weights at birth at the 5% level of significance? Select one: a. Fail to reject the null hypothesis. b. Reject the null hypothesis and conclude the mean is higher than 6.6 lb. c. Reject the null hypothesis and conclude the mean is lower than 6.6 lb. d. Cannot calculate because the population standard deviation is unknown.

Answer 1

Answer:

$t=\frac{7.557-6.6}{\frac{1.177}{\sqrt{7}}}=2.151$  

$p_v =2*P(t_{6}>2.151)=0.075$  

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 FAIL to reject the null hypothesis, we don't have enough evidence to conclude that the true mean is different from 6.6 pounds.

Based on this the best conclusion is:

a. Fail to reject the null hypothesis.

Step-by-step explanation:

Data given and notation

Data: 9.0, 7.3, 6.0, 8.8, 6.8, 8.4, 6.6

We can calculate the sample mean and standard deviation with these formulas:

$\bar X =\frac{\sum_{i=1}^n X_i}{n}$

$s =\sqrt{\frac{\sum_{i=1}^n (X_i -\bar X)^2}{n-1}}$

The results obtained are given below:  

$\bar X=7.557$ represent the sample mean  

$s=1.177$ represent the standard deviation for the sample

$n=7$ sample size  

$\mu_o =6.6$ 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 6.6, the system of hypothesis would be:  

Null hypothesis:$\mu = 6.6$  

Alternative hypothesis:$\mu \neq 6.6$  

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:  

$t=\frac{7.557-6.6}{\frac{1.177}{\sqrt{7}}}=2.151$  

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

$df=n-1=7-1=6$

What do you conclude?  

a. Use the critical value approach.

Assuming 95% of confidence and $\alpha=0.05$ we can use the t distribution with 6 degrees of freedom in order to calculate a critical value that accumulates 0.025 of the area on the tails of the distribution. We can use excel and the code to do this is given by: "=T.INV(1-0.025,6)". And we got the critical value $t_{\alpha/2}=2.446$.

Since our calculates value < critical value. We fail to reject the null hypothesis, and we can say that at 5% of significance, we don't have enough evidence to conclude that the true mean is different from 6.6 pounds.

b. Use the p-value approach

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

$p_v =2*P(t_{6}>2.151)=0.075$  

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 FAIL to reject the null hypothesis, we don't have enough evidence to conclude that the true mean is different from 6.6 pounds.

Based on this the best conclusion is:

a. Fail to reject the null hypothesis.