Question

A method currently used by doctors to screen women for possible breast cancer fails to detect cancer in 15% of women who actually have the disease. A new method has been developed that researchers hope will be able to detect cancer more accurately. A random sample of 70 women known to have breast cancer were screened using the new method. Of these, the new method failed to detect cancer in 8. Calculate the test statistic used by the researchers for the corresponding test of hypothesis.

Answer 1

`z=\frac{\hat p -p_o}{\sqrt{(p_o (1-p_o))/(n)}}` (1)

`z=\frac{0.114 -0.15}{\sqrt{(0.15(1-0.15))/(70)}}=-0.844`

Explanation:

Data given and notation

n=70 represent the random sample taken

X=8 represent the number of the new method failed to detect cancer

`\hat p=(8)/(70)=0.114` estimated proportion of number of the new method failed to detect cancer

`p_o=0.15` is the value that we want to test

`\alpha` represent the significance level

z would represent the statistic (variable of interest)

`p_v` represent the p value (variable of interest)

Concepts and formulas to use

We need to conduct a hypothesis in order to test the claim that the new method is able to detect cancer more accurately than the currently method (that means a lower rate of the proportion of interest) .:

Null hypothesis:
`p \geq 0.15`

Alternative hypothesis:
`p < 0.15`

When we conduct a proportion test we need to use the z statistic, and the is given by:

`z=\frac{\hat p -p_o}{\sqrt{(p_o (1-p_o))/(n)}}` (1)

The One-Sample Proportion Test is used to assess whether a population proportion
`\hat p` is significantly different from a hypothesized value
`p_o`.

Calculate the statistic

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

`z=\frac{0.114 -0.15}{\sqrt{(0.15(1-0.15))/(70)}}=-0.844`

Answer 2

The statistic is z=-0.674.

The null hypothesis failed to be rejected.

Explanation:

To make conclusions about the effectiveness of the new method, they should perform an hypothesis test on the proportion of failed cancer detection.

The actual method has a proportion of failed cancer detection of p=0.15. If the new method is better, it should have enough evidence that its actual proportion is below 0.15. This claim, that the new method proportion is below 0.15, will be stated in the alternative hypothesis.

The null and alternative hypothesis are:

`H_0: \pi=0.15\\\\H_a:\pi<0.15`

The sample size is n=70.

The sample proportion is p=8/70=0.114.

The standard error is calculated as:

`\sigma_p=\sqrt{(\pi(1-\pi))/(n)}=\sqrt{(0.15*0.85)/(70)}=√(0.0018)=0.043`

Then, the z-statistic for this sample is:

`z=(p-\pi+0.5/n)/(\sigma_p)=(0.114-0.15+0.5/70)/(0.043)=(-0.029)/(0.043) =-0.674`

The P-value for this statistic is:

`P-value=P(z<-0.674)=0.25`

At a significance level of 0.1, the P-value is bigger, so the effect is not significant. The null hypothesis failed to be rejected.