Statistical Hypothesis Testing Problems with Step-by-Step Solutions

This assignment provides detailed solutions to two hypothesis testing problems. The first problem involves testing whether more than 50% of Sydney people wanted tolls reduced, using a z-test and determining statistical significance based on the calculated z-value and p-value. The null hypothesis is rejected at a 5% level of significance. The second problem involves a t-test to determine if the average refrigerator energy usage is significantly different from a given value, including calculating a 95% confidence interval and assessing whether a particular energy usage value is unusual within the sample. The t-calculated value was within the acceptance region of 95%.

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Hypothesis Testing Problems
Student Name: Student ID:
Unit Name: Unit ID:
Date Due: Professor Name:

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Answer 1:
A. Let the null hypothesis be H0 : ( p0=0. 5 ) which is to be tested against the alternate
hypothesis H A : ( p≠0 . 5 ) .
From sample p
^¿=562/ 1040=0 .54
¿
Hence the choice of test is the z-test where
z= p
^¿−p o
√ po∗( 1− po )
n
= 0 . 54−0. 5
√ 0. 5∗0 .5
1040
= 0. 04
√ 0. 00024 = 0 . 04
0 . 0155 =2 .58 ¿
Now at 5% level of significance, the critical z-value is 1.96.
The z-calculated = 2.58 > z-critical = 1.96, hence the calculated z falls in the critical region.
Therefore, the null hypothesis is rejected at 5% level of significance. It can be concluded that
more than 50% of Sydney people wanted tolls reduced by 25 cents than lowering car registration
by $ 100.
B. As z-critical = 2.58, for the two tail z-test the p-value is the area under the normal curve
for |z|> 2. 58 . From standard normal table, P (|z| >2 .58 ) = 2∗0 .0049=0 . 0098
The p-value was less than 0.05. Hence, the z-calculated was statistically significant (Long,
Prober & Fischer, 2017)

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Answer 2:
A. Let the null hypothesis be H0 : ( μ0=439 ) which is to be tested against the alternate
hypothesis H A : ( μ>439 ) .
From sample, μA =445 , a sample SD = 40 with n = 18
Hence the choice of test is the t-test where
t= μ A−μ0
s / √ n−1 =445−439
40 / √ 17 = 6
9 .701 =0 . 62
Now at 5% level of significance, the critical t-value is for 17 degrees of freedom is 1.74 (from t-
table). Hence the t-calculated = 0.62 was less than t-critical = 1.74. Hence, the null hypothesis
got failed to be rejected at 5% level of significance.
B. The 95% confidence interval estimate of the population mean for t-test is ( x
−
− stc
√ n , x
−
+ stc
√ n )
where x-bar is the sample mean and s= √ n
n−1 S= √ 18
17∗40= √ 1 . 06∗40=41 .18 . S is the
sample SD.
Hence the 95% confidence interval estimate of the population mean is
( 445−41 . 18∗1 .74
√ 18 , 445+ 41. 18∗1 . 74
√ 18 )= ( 445−16 . 89 , 445+16 . 89 ) = ( 428 .11 , 461 . 89 )
. Incidentally, the
population mean was inside the 95% confidence interval.
C. For sample mean = 425 kWh / annum, the t-statistic will be as follows,
t = x
−
−μ0
s / √ n−1 =425−439
40 / √ 17 = −14
9 .701 =−1 . 44
Where t-critical = -1.74 for left tail test.
Hence, t-calculated was within the acceptance region of 95%. Hence, energy usage of 425 for a
particular refrigerator is not an unusual value in the sample. Though the value was outside the
confidence interval of part (b), the calculated p-value was 0.84 (using t-table) for 17 degrees of
freedom. Hence, the value was not an unusual value in the sample (Campbell & Stanley, 2015).

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Reference
Campbell, D.T. and Stanley, J.C., 2015. Experimental and quasi-experimental designs for
research. Ravenio Books.
Long, S.S., Prober, C.G. and Fischer, M., 2017. Principles and Practice of Pediatric Infectious
Diseases E-Book. Elsevier Health Sciences.