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Answer:
Step-by-step exhplanation:
a) n = 36 , M = 4.48 , = 4.22 , = 0.60 , = 0.05
The hypothesis are given by,
H0 : = 4.22 v/s H1 : 4.22
The Z statistic to test the hypothesis is given by,
Z = (M - μ) / (σ√n) ~ N(0,1)
Where the critical region /Z/ ≥ Zcritical for a two tailed test.
Z = (4.48 - 4.22)/(0.60√36)
Z = 2.6
Critical value = Z/2 = 1.96
The calculated value z > critical value z .
Thus, we reject the null hypothesis.
The number of fatty, sugary snacks eaten by overweight students is significantly different from the overall population mean.
b) n =25, M = 4.01 , = 4.22 , = 0.60 , = 0.05
The hypothesis is given by,
H0 : 4.22 v/s H1 : < 4.22
The Z statistic to test the hypothesis is given by,
Z = (M - μ) / (σ√n) ~ N(0,1)
Where the critical region /Z/ < Zcritical for a one tailed test.
Z = (4.01- 4.22)/(0.60√25)
Z = -1.75
The critical value of z = -1.645
The calculated value z > The critical value of z
Hence we reject the null hypothesis.
The healthy-weight students eat significantly fewer fatty, sugary snacks than the overall population.
The probability shows that since z = 2.6 which is more than the critical value of 1.96, the null hypothesis will be rejected.
How to calculate probability?
From the information given, the standard error will be:
= 0.60/✓36
= 0.60/6
= 0.10
The critical region will be:
= (4.48 - 4.22) / (0.60 × ✓36)
= 2.6
The critical value at 0.05 is 1.96.
Since z = 2.6 which is more than the critical value of 1.96, the null hypothesis will be rejected.
The students consume significantly different number of fatty, sugary drinks and snacks than the overall population average.
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