Question

A random sample of 180 microbiology students were asked how many science classes he or she was enrolled in August 1990. The results showed a mean of 1.83 science classes with a standard deviation of 1.48. Fifteen years later, a similar survey was conducted to determine if the distribution changed. The 2005 sample mean was 1.94 with a standard deviation of 1.62. Do the data provide statistical evidence that the mean number of science classes taken in the first survey is different from the survey taken 15 years later? Perform the appropriate test using α = 0.05.

Answer 1

Answer:

$z=\frac{1.83-1.94}{\sqrt{\frac{1.48^2}{180}+\frac{1.62^2}{180}}}}=-0.673$  

$p_v =2*P(z$

Comparing the p value with the significance level given $\alpha=0.05$ we see that $p_v>\alpha$ so we can conclude that we FAIL to reject the null hypothesis, and a would NOT be a significant difference in the two means

Step-by-step explanation:

Data given and notation

$\bar X_{1}=1.83$ represent the mean in 1990

$\bar X_{2}=1.94$ represent the mean for 2005

$s_{1}=1.48$ represent the sample deviation for 1990

$s_{2}=1.62$ represent the sample standard deviation for 2005

$n_{1}=180$ sample size for 1990

$n_{2}=180$ sample size for 2005

t would represent the statistic (variable of interest)

$\alpha=0.05$ significance level provided

Develop the null and alternative hypotheses for this study?

We need to conduct a hypothesis in order to check if the means for the two groups are different, the system of hypothesis would be:

Null hypothesis:$\mu_{1}=\mu_{2}$

Alternative hypothesis:$\mu_{1} \neq \mu_{2}$

Since we don't know the population deviations for each group, for this case is better apply a t test to compare means, and the statistic is given by:

$z=\frac{\bar X_{1}-\bar X_{2}}{\sqrt{\frac{s^2_{1}}{n_{1}}+\frac{s^2_{2}}{n_{2}}}}$ (1)

z-test: Is used to compare group means. Is one of the most common tests and is used to determine whether the means of two groups are equal to each other.

Calculate the value of the test statistic for this hypothesis testing.

Since we have all the values we can replace in formula (1) like this:

$z=\frac{1.83-1.94}{\sqrt{\frac{1.48^2}{180}+\frac{1.62^2}{180}}}}=-0.673$  

What is the p-value for this hypothesis test?

Since is a bilateral test the p value would be:

$p_v =2*P(z$

Based on the p-value, what is your conclusion?

Comparing the p value with the significance level given $\alpha=0.05$ we see that $p_v>\alpha$ so we can conclude that we FAIL to reject the null hypothesis, and a would NOT be a significant difference in the two means