Question

Summary statistics of birth weights of infants (in lbs.) born to mothers who are smokers and non-smokers are shown:

Group	n	Mean (lbs.)	Standard Dev (s)
X1 (Non-smokers)	11	7.4	1.1
X2 (Smokers)	14	6.1	0.8

(2-pts) Is this a Single Sample Z-test, a Paired t-test or a Two Independent Samples t-test?

(Circle one) Single Sample Z-test   /   Paired t-test   /   Two Independent Samples t-test

B.    (14-pts) Test the mean difference for significance: u₁ – u₂ = 0

(2-pts) Hypothesis statements (you choose either 1-sided or 2-sided test)

Ho: u₁ – u₂ ____________

Ha: u₁ – u₂ ____________

(2-pts) ₁ - ₂ = _______

(2-pts) SE = __________   note: (1.1)²/11 = 0.11 and (0.8)²/14 = 0.046

(1-pt) df_conservative = _________

(2-pts) Test statistic = __________

(2-pts) p-value approximation

(3-pts) Conclusion of significance

Answer 1

A. Two independent Samples t-test

Since mothers who are smokers are independent from mothers who are non-smokers, therefore we would use a two independent sample t-test.

B.

Let μ₁ = Mean weight of infants born to mothers who are non-smokers

μ₂ = Mean weight of infants born to mothers who are smokers

The null hypothesis is given by:

H₀ : μ₁ = μ₂  i.e. the mean weight of infants born to mothers who are non-smokers is same as the mean weigth of infants born to mothers who are smokers

while the alternative hypothesis is given by:

H_a : μ₁ > μ₂ i.e. the mean weight of infants born to mothers who are non-smokers is more than the mean weigth of infants born to mothers who are smokers

We are using a one tailed test

Test Statistic

Under H₀the test statistic is given by:

Where N₁ = first sample size

x̅₁= first sample mean

s₁² = first sample standard deviation

N₂ = second sample size

x̅₂= second sample mean

s₂² = second sample standard deviation

Now

• x̅₁ - x̅₂ = 7.4 - 6.1 = 1.3
• Also the pooled standard deviation is given by:

= ((10*1.1² + 13 * 0.8²)/(11+14-2))^0.5

= 0.942245

• Now Standard Error

= 0.942245 * (1/11+1/14)^0.5

= 0.37964

• degree of freedom = n1 + n2 - 2 = 11 + 14 - 2 = 2
• test statistic = 1.3 / 0.37964 = 3.424
• p-value = P(t₂₃ > 3.424) = 1 - 0.9988 = 0.0012

Conclusion of significance:

If the p-value = 0.0012 <= α ,reject the null hypothesis in favor of the alternative hypothesis. If the p-value > α, do not reject the null hypothesis.

Here even at even α = 1%, 0.0012<0.01, we reject the null hypothesis.

Therefore there is strong evidence against the null hypothesis.

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