Question

We want to test whether or not more students are absent on Friday afternoon classes than on Wednesday afternoon classes. In a random sample of 302 students with Friday afternoon classes, 48 missed the class. In a different random sample of 307 students with Wednesday afternoon classes, 32 missed the class. The table below summarizes this information. The standard error (SE) is given to save calculation time if you are not using software.

Data Summary

	total number	total number	Proportion
Day	of absences (x)	of students (n)	p̂ = (x/n)
Friday	48	302	0.15894
Wednesday	32	307	0.10423

SE = 0.02738

The Test: Test the claim that the absentee rate on all Friday afternoon classes is greater than the absentee rate on all Wednesday afternoon classes. Use a 0.05 significance level.

(a) Letting p̂₁ be the absentee rate from the sample on Friday and p̂₂ be the rate from Wednesday, calculate the test statistic using software or the formulaz =

(p̂1 − p̂2) − δp

SE

where δ_p is the hypothesized difference in proportions from the null hypothesis and the standard error (SE) given with the data. Round your answer to 2 decimal places.
z =
To account for hand calculations -vs- software, your answer must be within 0.01 of the true answer.

(b) Use software or the z-table to get the P-value of the test statistic. Round to 4 decimal places.
P-value =

(c) What is the conclusion regarding the null hypothesis?

reject H₀

fail to reject H₀    

(d) Choose the appropriate concluding statement.

The data supports the claim that the absentee rate on all Friday afternoon classes is greater than the absentee rate on all Wednesday afternoon classes.

There is not enough data to support the claim that the absentee rate on all Friday afternoon classes is greater than the absentee rate on all Wednesday afternoon classes.    

We have proven that the absentee rate on all Friday afternoon classes is greater than the absentee rate on all Wednesday afternoon classes.

We reject the claim that the absentee rate on all Friday afternoon classes is greater than the absentee rate on all Wednesday afternoon classes.

Answer 1

a)

Test statistic
z = (p1cap - p2cap)/sqrt(pcap * (1-pcap) * (1/N1 + 1/N2))
z = (0.1589-0.1042)/sqrt(0.131*(1-0.131)*(1/302 + 1/307))
z = 2.00

b)

P-value = 0.0228

c)

reject H0

d)

We reject the claim that the absentee rate on all Friday afternoon classes is greater than the absentee rate on all Wednesday afternoon classes.