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ivolga24 [154]
2 years ago
12

Please, help me. Thanks. Subject: Hypothesis testing and margin of errors

Mathematics
1 answer:
STatiana [176]2 years ago
4 0

Answer:

See below

Step-by-step explanation:

<u>Problem 4</u>

<u />H_0:\mu=145\\H_a:\mu>145

The null hypothesis, H_0, comes from the fact that the nurse initially thinks that the average height of 7th graders is equal to 145 cm.

The alternative hypothesis, H_a, comes from the fact that the average height may be greater than 145 cm since the nurse measured the average height to be 147 cm from the random sample.

<u>Problem 5</u>

<u />H_0:\mu\geq273\\H_a:\mu

The null hypothesis, H_0, comes from the fact that the manufacturer initially thinks that the mean lifetime of a certain type of batteries is at least 273 hours.

The alternate hypothesis, H_a, comes from the fact that the sample taken may show that it's actually less given a sample mean of 270.5 hours.

<u>Problem 6</u>

The mass population data set would be most suitable because the average score of a test comes from all those who have taken the test while the sample is the class with 25 students. That sample represents the population of all those that have taken the test.

<u>Problem 7</u>

Because it would take a lot of time and money to check every single child in Africa, it would make the most sense to use the sample data set where your samples come from each village and those samples have the children from each village.

<u>Problem 8</u>

This is a population characteristic because the sample that Sarah took for the average number of children per family represents the entire population.

<u>Problem 9</u>

This is a sample characteristic because the size of the sample was clearly large, this was a study, and other claims made.

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The percentage of body fat of a random sample of 36 men aged 20 to 29 found a sample mean of 14.42. Find a 95% confidence interv
Rina8888 [55]

Answer:

14.42-1.96\frac{6.95}{\sqrt{36}}=12.150    

14.42+ 1.96\frac{6.95}{\sqrt{36}}=16.690    

So on this case the 95% confidence interval would be given by (12.150;16.690)    

Step-by-step explanation:

Previous concepts

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".

The margin of error is the range of values below and above the sample statistic in a confidence interval.

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

\bar X=14.42 represent the sample mean

\mu population mean (variable of interest)

\sigma=6.95 represent the population standard deviation

n=36 represent the sample size  

Solution to the problem

The confidence interval for the mean is given by the following formula:

\bar X \pm z_{\alpha/2}\frac{\sigma}{\sqrt{n}}   (1)

Since the Confidence is 0.95 or 95%, the value of \alpha=0.05 and \alpha/2 =0.025, and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-NORM.INV(0.025,0,1)".And we see that z_{\alpha/2}=1.96

Now we have everything in order to replace into formula (1):

14.42-1.96\frac{6.95}{\sqrt{36}}=12.150    

14.42+ 1.96\frac{6.95}{\sqrt{36}}=16.690    

So on this case the 95% confidence interval would be given by (12.150;16.690)    

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