A new weight-watching company, Weight Reducers International, advertises that those who join will lose an average of 10 pounds a

Question

3 months ago

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A new weight-watching company, Weight Reducers International, advertises that those who join will lose an average of 10 pounds a

fter 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

Mathematics

1 answer:

Inessa [12.5K] · Answer 1 · 2026-03-02T12:01:03+03:00

Answer:

The value calculated is Z = 2.53, which exceeds 1.96 at a significance level of 0.05

This leads to the rejection of the null hypothesis H₀

Thus, we accept the alternative hypothesis

This implies that individuals participating in Weight Reducers will lose more than 10 pounds

Step-by-step explanation:

Step(i):-

The size of the random sample 'n' = 50

The new weight-loss program, Weight Reducers International, claims that members will shed an average of 10 pounds within the initial two weeks, with a standard deviation of 2.8 pounds.

Population mean 'μ' = 10 pounds

Population standard deviation 'σ' = 2.8 pounds

Sample mean 'x⁻' = 9

Significance level ∝ = 0.05

Step(ii):-

Null hypothesis: H₀: μ < 10

Alternative hypothesis: H₁: μ > 10

Test statistic calculation

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

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

The determined Z value is 2.53

In this case, the critical value Z is 1.96 at the 0.05 significance level

Step(iii):-

Calculated Z value of 2.53 is greater than 1.96 at a significance level of 0.05

Consequently, we reject the null hypothesis H₀

We accept the alternative hypothesis

Conclusion:-

We determine that individuals who sign up for Weight Reducers will experience a weight loss exceeding 10 pounds