Elementary Statistics: Hypothesis Testing Assignment, Spring 2019

This document presents solutions to a homework assignment in elementary statistics, focusing on hypothesis testing. The solutions cover various problems from Tasks 9.2, 9.3, and 9.4, including questions related to mean weight of salmon fish, pumpkin diameter, driving time, telephone call length, car mileage, and time taken to complete online homework. Each solution includes the null and alternative hypotheses, the calculated p-value, the significance level, and a conclusion on whether to reject or fail to reject the null hypothesis. The assignment also addresses problems involving proportions and sample variances, with detailed explanations for each case, guiding the user to understand the concepts and applications of hypothesis testing in different scenarios.

ELEMENTARY STATISTICS
STUDENT ID:
[Pick the date]

TASK 9.2
Question 1
(a) Null and alternative hypothesis
H0 :𝜇≥ 23
Ha : μ<23
Mean weight of salmon fish x=21.1
Standard deviation s=6
Sample size n=46
The t value= x−μ
s
√ n
= 21.1−23
6
√ 46
=−2.1477
Degree of freedom = n-1=46-1 = 45
The p value ( lower tail ) =0.0186
As per the sign used in alternative hypothesis, it is a lower tailed hypothesis test.
(b)The p value is 0.0186 which is inferior than the given level of significance i.e. 0.06 and as
a result of this, null hypothesis will be rejected.
Yes! Reject Null hypothesis.
(c) The lowest level of significance for acceptance of alternative hypothesis is 2%.
Question 2
(a) Null and alternative hypothesis
H0 :𝜇¿ 34 cm
Ha : μ>34 cm
2

Mean diameter of pumpkin x=37 cm
Standard deviation s=11cm
Sample size n=26
The t value= x−μ
s
√ n
= 37−34
11
√ 26
=1.3906
Degree of freedom = n-1=26-1 = 25
The p value ( upper tail ) =0.1766
As per the sign used in alternative hypothesis, it is a right tailed hypothesis test.
(b)The p value is 0.1766 which is far higher than the level of significance i.e. 0.04 and
therefore, null hypothesis will not be rejected.
No! Cannot Reject Null hypothesis.
(c) The possibility of mean diameter being higher cannot be accepted at any of the
significance levels given.
Question 3
(a) Null and alternative hypothesis
H0 :𝜇¿ 20 min
Ha : μ>20 min
Mean driving time to reach to work x=24 min
Standard deviation s=8 min
Sample size n=31
3

The t value= x−μ
s
√ n
= 24−20
8
√ 31
=2.7839
Degree of freedom = n-1=31-1 = 30
The p value ( upper tail )=0.0092
As per the sign used in alternative hypothesis, it is a right tailed hypothesis test.
(b) Yes! we do reject the null hypothesis because the p value computed is lower than
significance level of 1%.
(c) The lowest level of significance for null hypothesis rejection is 1%.
Question 4
(a) Null and alternative hypothesis
H0 :𝜇¿ 10 min
Ha : μ>10 min
Mean length of telephone call x=14.409 min
Standard deviation s=7.950min
Sample size n=22
The t value= x−μ
s
√ n
= 14.409−10
7.950
√ 22
=2.6013
Degree of freedom = n-1=22-1 = 21
The p value ( upper tail )=0.0167
As per the sign used in alternative hypothesis, it is a right tailed hypothesis test.
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(b) Yes! we do reject the null hypothesis because the p value found is lesser than significance
level of 4%.
(c) The lowest possible level of significance for acceptance of alternative hypothesis is 2%.
Question 5
(a) Null and alternative hypothesis
H0 :𝜇¿ 33
Ha : μ>33
Mean mileage per gallon of car x=35 miles
Standard deviation s=5.6 miles
Sample size n=19
The t value= x−μ
s
√ n
= 35−33
5.6
√ 19
=1.5567
Degree of freedom = n-1=19-1 = 18
The p value ( upper tail )=0.1369
As per the sign used in alternative hypothesis, it is a right tailed hypothesis test.
(b) No! we cannot reject the null hypothesis because the p value exceeds the significance
level of 4%.
(c) The new is not preferred over the old model for any of the given values of level of
significance.
Question 6
(a) Null and alternative hypothesis
H0 :𝜇≥ 45
5

Ha : μ<45
Mean time taken to finish an online HW x=42min
Standard deviation s=7.5min
Sample size n=27
The t value= x−μ
s
√n
= 42−45
7.5
√27
=−2.0785
Degree of freedom = n-1=27-1 =26
The p value ( lower tail )=0.0477
As per the sign used in alternative hypothesis, it is a left tailed hypothesis test.
(b) No! we cannot reject the null hypothesis because the p value exceeds the significance
level of 4%.
(c) The lowest level of significance for acceptance of alternative hypothesis 5%.
TASK 9.3
Question 1
(a) Null and alternative hypothesis
H0 : ^P=0.5
Ha : ^P>0.5
Proportion of population rise p= 119
211 =0.5640
Sample size n=211
The z value= p− ^P
√ ^P ( 1− ^P )
n
= 0.5640−0.5
√ 0.5 ( 1−0.5 )
211
=1.8588
6

The p value ( upper tail )=0.0315
As per the sign used in alternative hypothesis, it is a right tailed hypothesis test.
(b) Yes! we reject the null hypothesis because the p value does not exceed the significance
level of 4%.
(c) The lowest level of significance for acceptance of alternative hypothesis is 4%.
Question 2
(a) Null and alternative hypothesis
H0 : ^P=0.2
Ha : ^P>0.2
Proportion of population rise p= 64
276 =0.2318
Sample size n=211
The z value= p− ^P
√ ^P ( 1− ^P )
n
= 0.2318−0.2
√ 0.2 ( 1−0.2 )
276
=1.3242
The p value ( upper tail )=0.00
As per the sign used in alternative hypothesis, it is a right tailed hypothesis test.
(b) Yes! we reject the null hypothesis because the p value does not exceed the significance
level of 4%.
(c) The lowest level of significance for acceptance of alternative hypothesis is 0.1%.
Question 3
7

(a) Null and alternative hypothesis
H0 : ^P=0.13
Ha : ^P≠ 0.13
Proportion of population rise p= 76
721 =0.1054
Sample size n=211
The z value= p− ^P
√ ^P ( 1− ^P )
n
= 0.1054−0.13
√ 0.13 ( 1−0.13 )
721
=−1.9634
The p value ( two tail )=0.0496
As per the sign used in alternative hypothesis, it is a two tailed hypothesis test.
(b) No! we fail to reject the null hypothesis because the p value does exceed the significance
level of 4%.
(c) The lowest level of significance level to accept that the new machine is better is 5%.
TASK 9.4
Question 1
(a) Null and alternative hypothesis
H0 :σ2=196 ounces
Ha :σ2 >196 ounces
Sample variance s2=324 ounces
Sample size n=27
Stat valueY =¿
8

Degree of freedom = 27-1 = 26
The p value ( Upper tail )=0.019408
As per the sign used in alternative hypothesis, it is a right tailed hypothesis test.
(b) Yes! we reject null hypothesis because the p value does not exceed the significance level
of 5%.
(c) The lowest level of significance for acceptance of alternative hypothesis is 2%.
Question 2
(a) Null and alternative hypothesis
H0 :σ2=5 K
Ha :σ2 ≠5 K
Sample variance s2=8.8 K
Sample size n=26
Stat valueY = ( n−1 ) s2
σ2 = ( 26−1 )∗8.8
5 =44
Degree of freedom = 26-1 = 25
The p value ( Two tail )=0.010845
As per the sign used in alternative hypothesis, it is a two tailed hypothesis test.
(b) Yes! we reject null hypothesis because the p value does not exceed the significance level
of 4%.
(c) The lowest level of significance for acceptance of alternative hypothesis is 2%.
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Question 3
(a) Null and alternative hypothesis
H0 :σ2=20000 KWH square
Ha :σ2 >20000 KWH square
Sample variance s2=32500 KWH square
Sample size n=19
Stat valueY = ( n−1 ) s2
σ2 = ( 19−1 )∗32500
20000 =29.25
Degree of freedom = 19-1=18
The p value ¿ ¿ ¿=0.045408
As per the sign used in alternative hypothesis, it is a two tailed hypothesis test.
(b) No! we fail to reject null hypothesis because the p value is more than the significance
level of 4%.
(c) Lowest level of significance level for acceptance of alternative hypothesis is 5%.
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