Question

A psychologist obtains a random sample of 22 mothers in the first trimester of their pregnancy. the mothers are asked to play classical music in the house at least 30 min each day until they give birth. after 5 years, the child is administered an IQ test. It is known that IQs are normally distributed with a mean of 100. if the IQs of the 22 children in the study result in a sample mean of 105.7 ans sample standard deviation of 13, is there evidence that the children have higher IQs? Use the alphaαequals=0.050.05 level of significance. Complete parts​ (a) through​ (d).

​(a) Determine the null and alternative hypotheses.
(b.)Identify the test statistic.

i. Approximate the​ P-value.
ii. The​ P-value is in the range

​(c) State the conclusion for the test.
(d) State the conclusion in context of the problem.

Answer 1

Answer:

a) Null hypothesis:$\mu \leq 100$  

Alternative hypothesis:$\mu > 100$  

b)  $t=\frac{105.7-100}{\frac{13}{\sqrt{22}}}=2.057$  

c) $p_v =P(t_{(21)}>2.057)=0.0256$  

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.  

We can say that at 5% of significance the true mean is significant higher than 100.

d) On this case since we reject the null hypothesis. We can say that mean of the IQs for the children from the mothers that play classical music in the house is significant higher than 100.  

Step-by-step explanation:

Data given and notation  

$\bar X=105.7$ represent the sample mean  

$s=13$ represent the sample standard deviation  

$n=22$ sample size  

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

$\alpha=0.05$ 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)  

Part a. State the null and alternative hypotheses.  

We need to conduct a hypothesis in order to check if the population mean is higher than 100, the system of hypothesis are :  

Null hypothesis:$\mu \leq 100$  

Alternative hypothesis:$\mu > 100$  

Since we don't know the population deviation, 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".  

Part b. Calculate the statistic  

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

$t=\frac{105.7-100}{\frac{13}{\sqrt{22}}}=2.057$  

P-value  

We need to calculate the degrees of freedom first given by:  

$df=n-1=22-1=21$  

Since is a one-side right tailed test the p value would given by:  

$p_v =P(t_{(21)}>2.057)=0.0256$  

Part c. 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.  

We can say that at 5% of significance the true mean is significant higher than 100.

Part d. State the conclusion in context of the problem.

On this case since we reject the null hypothesis. We can say that mean of the IQs for the children from the mothers that play classical music in the house is significant higher than 100.