Respuesta :
Answer:
The calculated value Z = 2.53 >1.96 at 0.05 level of significance
Null hypothesis H₀ is rejected
we accepted alternative hypothesis
we conclude that those joining Weight Reducers will lose greater than 10 pounds
Step-by-step explanation:
Step(i):-
Given the random sample size 'n' =50
Given data a new weight-watching company, Weight Reducers International, advertises that those who join will lose an average of 10 pounds after the first two weeks. The standard deviation is 2.8 pounds.
The mean of the Population 'μ' = 10pounds
The standard deviation of the Population 'σ' = 2.8 pounds
Given mean of the sample 'x⁻' = 9
Level of significance ∝ = 0.05
Step(ii):-
Null hypothesis :H₀: μ<10
Alternative hypothesis: H₁: μ>10
The test statistic
[tex]z= \frac{x^{-} - mean}{\frac{S.D}{\sqrt{n} } }[/tex]
[tex]z= \frac{9 - 10}{\frac{2.8}{\sqrt{50} } } = \frac{-1}{0.395} = 2.53[/tex]
The calculated value Z = 2.53
The tabulated value Z = 1.96 at 0.05 level of significance
Step(iii):-
The calculated value Z = 2.53 >1.96 at 0.05 level of significance
Null hypothesis H₀ is rejected
we accepted alternative hypothesis
Conclusion:-
we conclude that those joining Weight Reducers will lose greater than 10 pounds
