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Hoochie [10]
3 years ago
5

A researcher took a random sample of 100 students from a large university. She computed a 95% confidence interval to estimate th

e average weight of the students at this university. The confidence interval was too wide to provide a precise estimate.True or false? The researcher could produce a narrower confidence interval by increasing the sample size to 150.A medical researcher is investigating the effect of drinking coffee on systolic blood pressure. The researcher assumes the average systolic blood pressure is 120 mmHg. For a random sample of 200 patients, the researcher takes two measurements of systolic blood pressure. The first systolic blood pressure measurement is taken during a week when the patients drink no coffee, and the second systolic blood pressure measurement is taken during a week when the patients drink at least two cups of coffee. The medical researcher wonders whether there is a significant difference between the blood pressure measurements. Which of the following is the correct null and alternative hypothesis for the medical researchers study?
Mathematics
1 answer:
katen-ka-za [31]3 years ago
7 0

Answer:

Part (A): The correct option is true.

Part (B): The null and alternative hypothesis should be:

H_o: \mu =0 ; H_a:\mu \neq0

Step-by-step explanation:

Consider the provided information.

Part (A)

A random sample of 100 students from a large university.

Increasing the sample size decreases the confidence intervals, as it increases the standard error.

If the researcher increase the sample size to 150 which is greater than 100 that will decrease the confidence intervals or the researcher could produce a narrower confidence interval.

Hence, the correct option is true.

Part (B)

The researcher wants to identify that whether there is any significant difference between the measurement of the blood pressure.

Therefore, the null and alternative hypothesis should be:

H_o: \mu =0 ; H_a:\mu \neq0

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\displaystyle \int_{x=0}^{x=\frac\pi2} \tan(x) \ln(\sin(x)) \, dx = \int_{y=\sin(0)}^{y=\sin\left(\frac\pi2\right)} \frac{y}{\sqrt{1-y^2}} \ln(y) \frac{dy}{\sqrt{1-y^2}}

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For 0 < y < 1, we have |1 - y²| = 1 - y², so

\displaystyle \int_0^1 \frac{y}{1-y^2} \ln(y) \, dy = uv \bigg|_{y\to0^+}^{y\to1^-} + \frac12 \int_0^1 \frac{\ln(1-y^2)}{y} \, dy

It's easy to show that uv approaches 0 as y approaches either 0 or 1, so we just have

\displaystyle \int_0^1 \frac{y}{1-y^2} \ln(y) \, dy = \frac12 \int_0^1 \frac{\ln(1-y^2)}{y} \, dy

Recall the Taylor series for ln(1 + y),

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