Question

A sample of 12 radon detectors of a certain type was selected, and each was exposed to 100 pCI/L of radon. The resulting readings were as follows: 105.6, 90.9, 91.2, 96.9, 96.5, 91.3, 100.1, 105.0, 99.6, 107.7, 103.3, 92.4 Does this data suggest that the population mean reading under these conditions differs from 100? Set up an appropriate hypothesis test to answer this question.

Answer 1

Answer:

Null hypothesis:$\mu = 100$  

Alternative hypothesis:$\mu \neq 100$  

$t=\frac{98.375-100}{\frac{6.109}{\sqrt{12}}}=-0.921$  

$p_v =2*P(t_{11}<-0.921)=0.377$  

Step-by-step explanation:

1) Data given and notation  

Data: 105.6, 90.9, 91.2, 96.9, 96.5, 91.3, 100.1, 105.0, 99.6, 107.7, 103.3, 92.4

We can calculate the sample mean and deviation for this data with the following formulas:

$\bar X =\frac{\sum_{i=1}^n X_i}{n}$

$s=\sqrt{\frac{\sum_{i=1}^n (X_i- \bar X)^2}{n-1}}$

The results obtained are:

$\bar X=98.375$ represent the sample mean  

$s=0.6.109$ represent the sample standard deviation  

$n=12$ sample size  

$\mu_o =100$ represent the value that we want to test  

$\alpha$ represent the significance level for the hypothesis test.  

z would represent the statistic (variable of interest)  

$p_v$ represent the p value for the test (variable of interest)  

2) State the null and alternative hypotheses.  

We need to conduct a hypothesis in order to check if the mean is equal to 100pCL/L, the system of hypothesis are :  

Null hypothesis:$\mu = 100$  

Alternative hypothesis:$\mu \neq 100$  

Since we don't know the population deviation, is better apply a t test to compare the actual mean to the reference value, and the statistic is given by:  

$t=\frac{\bar X-\mu_o}{\frac{s}{\sqrt{n}}}$ (1)  

t-test: "Is used to compare group means. Is one of the most common tests and is used to determine if the mean is (higher, less or not equal) to an specified value".  

3) Calculate the statistic  

We can replace in formula (1) the info given like this:  

$t=\frac{98.375-100}{\frac{6.109}{\sqrt{12}}}=-0.921$  

4) P-value  

First we need to find the degrees of freedom for the statistic given by:

$df=n-1=12-1=11$

Since is a two sided test the p value would given by:  

$p_v =2*P(t_{11}<-0.921)=0.377$  

5) Conclusion  

If we compare the p value and the significance level assumed $\alpha=0.05$ we see that $p_v>\alpha$ so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, so we can conclude that the true mean is not significant different from 100 at 5% of significance.