America Patton

2022-03-16

Calculating the Confidence Interval

It is found that in a random sample of 100 Science students, there are 48 studying statistics . To test whether the true proportion of students in statistics is 50% or not, suitable null and alternative hypotheses are:

(a)$H0:p=0.48;H1:p\u0338=0.48$

(b)$H0:p=0.5;H1:p>0.5$

(c)$H0:p=0.5;H1:p<0.5$

(d)$H0:p=0.5;H1:p=0.48$

(e)$H0:p=0.5;H1:p\u0338=0.5$

Find a 95% Confidence Interval for p in Q1.

I have determined the answer to Q

1), which I believe is e).

For question 2), I have been told the CI= estimator +/- (table value)(Standard error). The estimator is 0.48, and I have found the SE to be 0.05. However, I am unsure about how to calculate the "table value".

It is found that in a random sample of 100 Science students, there are 48 studying statistics . To test whether the true proportion of students in statistics is 50% or not, suitable null and alternative hypotheses are:

(a)

(b)

(c)

(d)

(e)

Find a 95% Confidence Interval for p in Q1.

I have determined the answer to Q

1), which I believe is e).

For question 2), I have been told the CI= estimator +/- (table value)(Standard error). The estimator is 0.48, and I have found the SE to be 0.05. However, I am unsure about how to calculate the "table value".

Veronica Riddle

Beginner2022-03-17Added 9 answers

You are correct to use ${H}_{0}:p=.5$ versus ${H}_{1}:p\ne 0.5$ . for your hypothesis testing.

In the US until several years ago, the traditional 95% ci for p is to use the point estimate$\hat{p}=\frac{X}{n}=\frac{48}{100}$ . The standard error of $\hat{p}$ is $\sqrt{p\frac{1-p}{n}}$ , but because you don't know p, the traditional method is to estimate it by plugging in $\hat{p}$ for p. Thus the traditional formula for the 95% CI is

$\hat{p}\pm 1.96\sqrt{\frac{\hat{p}(1-\hat{p})}{n}},$

which computes to (0.382,0.578). This method uses the normal approximation to the binomial distribution, so 1.96 is the 'table value' you're looking for.

However, this method has two sources of error: (a) using the normal approximation to the binomial distribution, and (b) estimating the standard error. Computational evidence is that the traditional method often does not actually have the 'promised' 95% coverage.

Much of the error due to (b) can be eliminated by using$\stackrel{~}{n}=n+2$ instead of n, and $\stackrel{~}{p}=\frac{X+2}{\stackrel{~}{n}}=\frac{50}{104}$ instead of $\hat{p}$ . The resulting adjusted interval, due to Agresti, is (0.385,0.577). (In your particular example, the adjustment is small.)

The Wilson interval is also more accurate than the traditional one. You can google that. For the 95% confidence level, it is not much different from the more easily computed Agresti interval. The Wilson interval has been standard for many years in parts of Europe.

All three intervals cover$p=0.5$ , so there is no reason to doubt that the true percentage of statistics majors is near 50%.

In the US until several years ago, the traditional 95% ci for p is to use the point estimate

which computes to (0.382,0.578). This method uses the normal approximation to the binomial distribution, so 1.96 is the 'table value' you're looking for.

However, this method has two sources of error: (a) using the normal approximation to the binomial distribution, and (b) estimating the standard error. Computational evidence is that the traditional method often does not actually have the 'promised' 95% coverage.

Much of the error due to (b) can be eliminated by using

The Wilson interval is also more accurate than the traditional one. You can google that. For the 95% confidence level, it is not much different from the more easily computed Agresti interval. The Wilson interval has been standard for many years in parts of Europe.

All three intervals cover

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