Question

Researchers obtained a sample of n = 10 children who attended daycare before starting school. The children are given a standardized math test for which the population mean is µ = 50. The scores for the sample are as follows: 53, 57, 61, 49, 52, 56, 58, 62, 51, 56. Is this sample sufficient to conclude that the children with a history of preschool daycare are significantly different from the general population?

Answer 1

Answer:

$t=\frac{55.5-50}{\frac{4.249}{\sqrt{10}}}=4.093$  

$p_v =2*P(t_{9}>4.093)=0.00084$  

If we compare the p value and the significance level assumed $\alpha=0.05$ we see that $p_v$ 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 significantly different from 50.

Step-by-step explanation:

Data given and notation  

53, 57, 61, 49, 52, 56, 58, 62, 51, 56

We can calculate the sample mean and deviation with the following formulas:

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

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

And we got:

$\bar X=55.5$ represent the sample mean  

$s=4.249$ represent the sample standard deviation  

$n=10$ sample size  

$\mu_o =50/tex] represent the value that we want to test [tex]\alpha=0.05$ represent the significance level for the hypothesis test (ASSUMED).  

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.  

We need to conduct a hypothesis in order to check if the true mean is equal or not to 50 :  

Null hypothesis:$\mu = 50$  

Alternative hypothesis:$\mu \neq 50$  

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".  

Calculate the statistic  

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

$t=\frac{55.5-50}{\frac{4.249}{\sqrt{10}}}=4.093$  

P-value  

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

$df=n-1=10-1=9$  

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

$p_v =2*P(t_{9}>4.093)=0.00084$  

Conclusion  

If we compare the p value and the significance level assumed $\alpha=0.05$ we see that $p_v$ 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 significantly different from 50.