Question

Given the sample mean = 21.15, sample standard deviation = 4.7152, and N = 40 for the low income group,

Test the claim that the mean nickel diameter drawn by children in the low income group is greater than 21.21 mm. Test at the 0.1 significance level.

a) Identify the correct alternative hypothesis:

• p=21.21p=21.21
• μ>21.21μ>21.21
• μ=21.21μ=21.21
• μ<21.21μ<21.21
• p<21.21p<21.21
• p>21.21p>21.21

Give all answers correct to 3 decimal places.

b) The test statistic value is:     

c) Using the Traditional method, the critical value is:  

d) Based on this, we

• Reject H0H0
• Fail to reject H0H0

e) Which means

• The sample data supports the claim
• There is sufficient evidence to warrant rejection of the claim
• There is not sufficient evidence to support the claim
• There is not sufficient evidence to warrant rejection of the claim

Answer 1

From the given information,

n = 40

Xbar = 21.15

s = 4.7152

α = 0.10

1. Null hypothesis H₀ : µ = 21.21

Alternative Hypothesis H_a : µ > 21.21

2. Test statistics (t) = (Xbar - μ) / (S/ √n)

t = (21.15 – 21.21)/ (4.7152/sqrt(40))

Test statistics t = -0.080

3. The one-sided t critical value at degrees of freedom n-1 = 39 and α=0.10

T_{39, 0.10} = 1.304 (from statistical table)

Rejection region: Reject H₀ if test statistics t > 1.304.

4. Since t statistics = -0.080 < t_{39, 0.10} = 1.304 that means t statistics value not fall in rejection region. So, we failed to reject H₀ at 10% level of significance.

5. Conclusion: We can conclude that, there is not sufficient evidence to support the claim that the mean nickel diameter drawn by children in the low-income group is greater than 21.21 mm at 0.10 level of significance.