Answer to Question #202606 in Statistics and Probability for Stephen Galicia

Question

 A company produced ethyl alcohol and claimed to have a mean alcohol content of

70%. A random sample of 80 of ethyl alcohol was take as sample to verify this claim. It

was found out that the mean alcohol content is 65% with a standard deviation of 2%. Test

the claim at 5% level of significance and assume that the population is normally

distributed​

Answer 1

Given claim

The mean alcohol content in ethyl alcohol produced by company is of 70%.

The null and alternate hypothesis are shown below

$H_0: μ=70 \\ H_a: μ≠70$

The sample mean alcohol content is \bar{X}=65% and standard deviation is σ=2%

The level of significance is 5%.

The population is normally distributed and as the sample size is high the sample is also treated as normally distributed.

The test statistic is calculated as shown below

$Z= \frac{\bar{X} \mu}{\sigma} \\ Z= \frac{65-70}{2}=-2.5$

This is a two tailed hypothesis test.

P-value $=2 \times P(Z<-2.5)$

With respect to Z table we get P(Z<-2.5)=0.0062

So p value is calculated as shown below

p-value $=2 \times P(Z< -2.5)=2 \times 0.0062=0.0124$

The p-value 0.0124 is less than the level of significance 0.05 or 5%. Thus we have sufficient evidence to reject the null hypothesis. So we conclude that the alcohol content in ethyl alcohol produced by company is not 70%.