Question

**Absentee rates - Friday vs Wednesday:** 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, 30 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 | 30 | 307 | 0.09772 |

*SE* = 0.02708

**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 p̂_{1} be the absentee rate from the sample on
Friday and p̂_{2} be the rate from Wednesday, calculate the
test statistic using software or the formulaz =

(p̂_{1} − p̂_{2}) − δ_{p} |

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 *H*_{0}

fail to reject
*H*_{0}

(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

Ans a ) using minitab stat>basic stat>two sample proportion

we have

Test and CI for Two Proportions

Sample X N Sample p

1 48 302 0.158940

2 30 307 0.097720

Difference = p (1) - p (2)

Estimate for difference: 0.0612205

95% lower bound for difference: 0.0167838

Test for difference = 0 (vs > 0): Z = 2.26 P-Value = 0.0119

Ans a ) p̂_{1} be the absentee rate from the
sample on Friday and p̂_{2} be the rate from Wednesday

from the output , (p̂_{1} − p̂_{2}) = 0.06

standard error (SE) = 0.03

the test statistic z = 2.36

b ) p value is 0.0119

c ) since p value is less than 0.05 so reject H0.

d )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.

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to test whether or not more students are absent on Friday afternoon
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302 students with Friday afternoon classes, 48 missed the class. In
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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.
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