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2 years ago

Please do part b A=5 and B=4

Mathematics

2 answers:

tatyana61 [14]2 years ago

4 0

Answer:

you have the answer already

Step-by-step explanation:

Scilla [17]2 years ago

3 0

Rectangle is 72

The square is 45

The length for A is 8

The length for B is 3

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

cupoosta [38]

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:

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

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

6 0

3 years ago

What percent of a circle is 3/6 of a circle

Vlad1618 [11]

50%
You have half of 6 so the percentage would be half of 100

8 0

3 years ago

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