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masha68 [24]
2 years ago
7

A recent health report revealed that a woman with insurance spends an average of 2.3 days in the hospital following a routine ch

ildbirth, while a woman without insurance spends an average of 1.9 days at the hospital. Two samples of 16 women each were used in both samples. The standard deviation of the first sample is equal to 0.6 day, and the standard deviation of the second sample is 0.3 day. At ???? =0.01, test the claim that the means are equal. Find the 99% confidence interval for the differences of the means and compare the result of the hypothesis test to the one for the confidence interval
Mathematics
1 answer:
cupoosta [38]2 years ago
6 0

Answer:

t=\frac{2.3-1.9}{\sqrt{\frac{0.6^2}{16}+\frac{0.3^2}{16}}}}=2.385  

p_v =2*P(t_{30}>2.385)=0.0236

Comparing the p value with the significance level given \alpha=0.01 we see that p_v>\alpha so we can conclude that we FAIL to reject the null hypothesis

The confidence interval would be given by:

(2.3 -1.9) - 2.75*\sqrt{\frac{0.6^2}{16}+\frac{0.3^2}{16}}}=-0.0611

(2.3 -1.9) + 2.75* \sqrt{\frac{0.6^2}{16}+\frac{0.3^2}{16}}}=0.861

Step-by-step explanation:

Data given and notation

\bar X_{insurance}=2.3 represent the mean for insurance

\bar X_{No. ins}=1.9 represent the mean withour insurance

s_{Insurance}=0.6 represent the sample standard deviation for the insurance case

s_{No. Ins}=0.3 represent the sample standard deviation for the No insurance case

n_{Insurance}=16 sample size for insurance

n_{No. Iss}=16 sample size for no insurance

t would represent the statistic (variable of interest)

\alpha=0.01 significance level provided

Develop the null and alternative hypotheses for this study?

We need to conduct a hypothesis in order to check if the means for the two groups are different, the system of hypothesis would be:

Null hypothesis:\mu_{Insu}=\mu_{No. Ins}

Alternative hypothesis:\mu_{Ins} \neq \mu_{No. Ins}

Since we know the population deviations for each group, for this case is better apply a z test to compare means, and the statistic is given by:

t=\frac{\bar X_{Ins}-\bar X_{No.Ins}}{\sqrt{\frac{s^2_{Ins}}{n_{Ins}}+\frac{s^2_{No. Ins}}{n_{No. Ins}}}} (1)

t-test: Is used to compare group means. Is one of the most common tests and is used to determine whether the means of two groups are equal to each other.

Calculate the value of the test statistic for this hypothesis testing.

Since we have all the values we can replace in formula (1) like this:

t=\frac{2.3-1.9}{\sqrt{\frac{0.6^2}{16}+\frac{0.3^2}{16}}}}=2.385  

What is the p-value for this hypothesis test?

The degrees of freedom are given by:

df = n_{ins} +n_{No Ins}-2=16+16-2 =30

Since is a bilateral test the p value would be:

p_v =2*P(t_{30}>2.385)=0.0236

Based on the p-value, what is your conclusion?

Comparing the p value with the significance level given \alpha=0.01 we see that p_v>\alpha so we can conclude that we FAIL to reject the null hypothesis

The confidence interval would be given by:

(2.3 -1.9) - 2.75*\sqrt{\frac{0.6^2}{16}+\frac{0.3^2}{16}}}=-0.0611

(2.3 -1.9) + 2.75* \sqrt{\frac{0.6^2}{16}+\frac{0.3^2}{16}}}=0.861

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