Question

Is there a difference between the distribution of community college statistics students and the distribution of university statistics students in what technology they use on their homework? Of some randomly selected community college students, 43 used a computer, 102 used a calculator with built in statistics functions, and 65 used a table from the textbook. Of some randomly selected university students, 28 used a computer, 33 used a calculator with built in statistics functions, and 40 used a table from the textbook. Conduct an appropriate hypothesis test using a 0.05 level of significance.

Read the statement and decide whether it is true or false.

Short answer

There is sufficient evidence to insure that the distribution of college statistics students is different from the distribution of university statistics students.

Step-by-step solution

Given Information

Given that randomly selected community college students, 43 used a computer, 102 used a calculator with built-in statistics functions, and 65 used a table from the textbook.

Of some randomly selected university students, 28 used a computer, 33 used a calculator with built-in statistics functions, and 40 used a table from a textbook.

Null Hypothesis

The null hypothesis is shown below: $H_0$: The distribution of college statistics students is same as the distribution of university statistics students.

Alternative Hypothesis

Against the alternative hypothesis as shown below:

$H_a:$The distribution of college statistics students is different from the distribution of university statistics students.

Degrees of Freedom

The degrees of freedom can be calculated by the formula given below:

$df=(numberofcolumn-1)$

Therefore

$df=3-1$

=2

$df=3-1$

Distribution

From above calculation, it is clear that the distribution for the test is ${\chi }_2^2$.

Expected  Frequencies

The observed value table is already given in the textbook. Calculate the expected frequencies by using the formula given below:

$E=\frac{\left(\text{row total}\right)\left(\text{column total}\right)}{\text{overall total}}$

All calculations can be done in excel worksheet. Hence, the expected localid="1649136146590" $\left(E\right)$values table is shown below:

Test statistic

The test statistic of the independence test is given below

$\text{Test statistic}=\sum _{(i\times f)}\frac{(O-E{)}^2}{E}$

To calculate $\frac{(O-E{)}^2}{E}$apply the formula

$=\left((B4-B11{)}^2\right)/B11$

same formula up to cell D19. After that, take the total of columns total and rows total. The table of the test statistics is shown below:

p-Value

The $p$-value can be calculated in excel by using CHIDIST ( ) formula as shown below:

Hence the p-value is 0..29

The Chi-square Sketch

The Chi-square sketch is given below

Final Answer

i. Alpha:$0.05$

ii: Decision: Reject the null hypothesis$H_0$

iii. Reason for decision: Because$p$-value $<\alpha$

iv. Conclusion: There is sufficient evidence to insure that the distribution of college statistics students is different from the distribution of university statistics students.