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A nutritionist compared weight loss success between two diets, A and B, using samples of 80 and 100 people respectively. The hypothesis test at a 0.05 significance level resulted in a calculated z-score of 0.34, which did not exceed the critical value of ±1.96, leading to a failure to reject the null hypothesis. This indicates no significant difference in weight loss success between the two diets.

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26 views2 pages

As 3

A nutritionist compared weight loss success between two diets, A and B, using samples of 80 and 100 people respectively. The hypothesis test at a 0.05 significance level resulted in a calculated z-score of 0.34, which did not exceed the critical value of ±1.96, leading to a failure to reject the null hypothesis. This indicates no significant difference in weight loss success between the two diets.

Uploaded by

wilbertconsul07
Copyright
© © All Rights Reserved
We take content rights seriously. If you suspect this is your content, claim it here.
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3.

A nutritionist wants to compare the proportion of people following two


different diets, Diet A and Diet B, who successfully lost weight. In a sample of 80
people following Diet A, 50 achieved weight loss. In a sample of 100 people
following Diet B, 60 achieved weight loss. The nutritionist will test if there is a
significant difference between the proportions at a 0.05 significance level.

Hypothesis:
H₀: There’s no significant difference in the proportions of weight loss success
between Diet A and Diet B. (p₁ = p₂)
H₁: There is significant difference in the proportions of weight loss success
between Diet A and Diet B. (p₁ ≠ p₂)
Level of significance: α=0.05
Test Statistic: Z-test

values are ±1.96. Since ∣0.34∣<1.96, we fail to reject the null hypothesis.
Critical Region: At a 0.05 significance level for a two-tailed test, the critical z-

Computation:
n1 = 80 x1 =50
n2 =100 x2 = 60

( p 1− p 2 )
z=

√ p 1(1− p 1) p 2(1− p 2)
n1
+
n2
( 0. 625−0.6 )
z=

√ 0. 625(1−0. 625) 0.6 (1−0. 6)


80
+
100
0. 025
z=

√ 0. 625( 0.375) 0. 6( 0. 4)
80
+
100
0. 025
z=

√ 0. 2344 0. 24
80
+
100
0.025
z=
√ 0.00293+ 0.00 24
0. 025
z=
0.0 730
z=0.34
Decision:
Since the calculated z-score (0.34) is less than the critical value of 1.96,
Therefore we fail to reject the null hypothesis.

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