Question

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. A random sample of 50 people who joined the weight reduction program revealed a mean loss of 9 pounds. At the 0.05 level of significance, can we conclude that those joining Weight Reducers will lose less than 10 pounds

Answer 1

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

$z= \frac{x^{-} - mean}{\frac{S.D}{\sqrt{n} } }$

$z= \frac{9 - 10}{\frac{2.8}{\sqrt{50} } } = \frac{-1}{0.395} = 2.53$

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