Question

A dietician believes that people who eat a high-fiber cereal as part of their breakfast will consume, on average, fewer calories at lunch than people who do not eat a high-fiber cereal as part of their breakfast. To test this claim, he measured the lunchtime calorie intake of 49 adults who did eat a high-fiber cereal for breakfast on a given day (group 1). For those individuals, the average number of calories consumed at lunch was 609.86, with a standard deviation of 55.96 calories. He also measured the lunchtime calorie intake of 78 adults who did not eat a high-fiber cereal for breakfast on a given day (group 2). For those individuals, the average number of calories consumed at lunch was 641.02, with a standard deviation of 109.14 calories. The dietician wishes to test this claim at the 5% significance level.

If Welch's two-sample test for equality of means is used, then what is the value of the test statistic tobs, rounded to two decimal places?

Answer 1

Answer:

The calculated value Z = 1.8368 < 1.96 at 0.05 level of significance

The null hypothesis is accepted

The samples have been drawn from the same Population

Step-by-step explanation:

Step(i):-

Given first sample size 'n₁' = 49

Mean of the first sample 'x₁⁻ = 609.86

Standard deviation of the sample S₁ = 55.96 calories

Given first sample size 'n₂' = 78

Mean of the first sample 'x₂⁻ = 641.02

Standard deviation of the sample S₂ = 109.14 calories

Step(ii):-

Null hypothesis : H₀: x₁⁻ = x₂⁻

Alternative Hypothesis : H₁: x₁⁻ ≠ x₂⁻

Level of significance ∝ = 0.05

Test statistic

$Z = \frac{x^{-} _{1}-x^{-} _{2} }{\sqrt{S.D^2(\frac{1}{n_{1} } }+\frac{1}{n_{2} } }$

where

 $S.D^{2} = \frac{n_{1} S^2_{1}+ n_{2} S^2_{2} }{n_{1} + n_{2} -2}$

 $= \frac{49 X (55.96)^2+ 78X(109.14)^2 }{49 + 78 }$

 σ² =  8660.357

Step(iii):-

Test statistic

$Z = \frac{x^{-} _{1}-x^{-} _{2} }{\sqrt{S.D^2(\frac{1}{n_{1} } }+\frac{1}{n_{2} } }$

$= \frac{ 609.86-641.02 }{\sqrt{8660.357(\frac{1}{49 } }+\frac{1}{78 } ) }$

$Z = \frac{-31.16}{16.9638} = -1.8368$

|Z| = |-1.8368| = 1.8368

The tabulated value Z₀.₉₅ = 1.96

The calculated value = 1.8368 < 1.96 at 0.05 level of significance

The null hypothesis is accepted

Final answer:-

The samples have been drawn from the same Population