Data from the latest census was analyzed. It was determined that one-parent households spend less on food away from home that do two-parent households. The mean amount spent by 30 one-parent households is $1876 and the standard deviation is $113. The mean amount spent by 30 two-parent households is $1878 and the standard deviation is $85. At a a=0.05 can you support this conclusion?

(please provide complete detailed answer)

To determine if the conclusion that one-parent households spend less on food away from home than two-parent households is supported by the data, we can conduct a hypothesis test.

First, let's establish the hypothesis statements:

Null Hypothesis (H0): The mean amount spent by one-parent households is equal to the mean amount spent by two-parent households. μ1 = μ2

Alternative Hypothesis (Ha): The mean amount spent by one-parent households is less than the mean amount spent by two-parent households. μ1 < μ2

Next, we need to determine the test statistic. Since we are comparing the means of two independent samples, we can use the independent samples t-test.

The formula for the test statistic is:
t = (x1 - x2) / sqrt((s1^2 / n1) + (s2^2 / n2))

Where:
x1 and x2 are the sample means,
s1 and s2 are the sample standard deviations,
n1 and n2 are the sample sizes.

Let's calculate the test statistic:

x1 = $1876
s1 = $113
n1 = 30

x2 = $1878
s2 = $85
n2 = 30

t = ($1876 - $1878) / sqrt(($113^2 / 30) + ($85^2 / 30))

Now, we need to find the critical value for our test at a significance level of α = 0.05. Since the alternative hypothesis is one-sided, we will look up the critical value in the t-distribution table using degrees of freedom equal to n1 + n2 - 2 (because we have two independent samples).

Degrees of freedom (df) = n1 + n2 - 2 = 30 + 30 - 2 = 58

Looking up the critical value in the t-distribution table for a one-sided test at α = 0.05 and df = 58, we find t_critical = -1.676.

Now, we can compare the test statistic t to the critical value t_critical to make a decision.

If t < t_critical, we reject the null hypothesis and support the alternative hypothesis.

If t ≥ t_critical, we fail to reject the null hypothesis.

Let's calculate the value of t and make a decision:

t = ($1876 - $1878) / sqrt(($113^2 / 30) + ($85^2 / 30))

After calculating t, we get the t-value.

Now, compare the calculated t-value to the critical value:

If t < t_critical, we reject the null hypothesis and support the alternative hypothesis.

If t ≥ t_critical, we fail to reject the null hypothesis.

If the decision is to reject the null hypothesis, it means that there is evidence to support the conclusion that one-parent households spend less on food away from home than two-parent households.

Please plug in the values and follow the steps to find the test statistic and make the decision.