In: Statistics and Probability
1.A fast-food restaurant selling soft drinks advertises that they contain 15 ounces. A manager wants to find out whether the drinks are being overfilled. A random sample of 10 independently chosen drinks produced a mean of 15.6 ounces with a standard deviation of 0.4 ounce. Assume that the distribution of serving sizes in the population is Normally distributed. Carry out the appropriate hypothesis test using a 1% significance level. Explain what conclusion the manager should reach.
2.The mean weight of all 21-year-old women is 132 pounds. A random sample of 35 vegetarian women who are 21 years old showed a sample mean of 127 pounds with a standard deviation of 15 pounds. The women's measurements were independent of each other. Determine whether the mean weight for 21-year old vegetarian women is significantly less than 132, using a significance level of 5%.
Question 1:
A fast food selling restaurant advertises that they serve soft drinks measuring 15 ounces.
The manager wishes to perform a quality control check whether the drinks are being overfilled or not.
We are given that a random sample of 10 drinks (=n say) gives a mean of 15.6 ounces (= say) with a standard devation of 0.4 ounce (= say).
Let X denote the quantity of drink being served .
By the question , X~ Normal (, )
To ensure that the test results help the manager make appropriate conclusions, we are to test the hypothesis :
versus
Note that the alternative hypothesis H1 is both sided since it would be of great concern to the manager if the drinks are being under served or over filled as well.
Since ,the population is given to be Normally distributed hence the sample elements drawn from it will follow Normal Distribution as well.
Let Xi denote the measurement of i-th sample of the drink where i=1,2,3,.........,10
Then , Xi ~ N (, )
then we know that ~ N(,) where is the mean of n sample elements
The test statistic that must be used to conduct the test on our hypothesis is called the t-Test.
t is defined as ,
where t follows a t-distribution with (n-1) degrees of freedom
i.e. t~tn-1
For our given data, the observed value of t is :
= 4.743
Now whether we should accept our null hypothesis or not is based on the camparison of our observed t value with the t value obtained from tables at 1% (=0.01) level of significance with 9 degrees of freedom (n-1=10-1).
From table, t0.01/2;9= 3.250 (We use 0.01/2 since our H1 is both sided)
Since our |observed t |> t0.01/2;9 , we reject our Ho and conclude that the soft drinks served are not being filled with 15 ounces at 1% level of significance.
Question 2 :
The mean weight of all 21-year-old women is 132 pounds. A random sample of 35 (=n say) vegetarian women who are 21 years old showed a sample mean of 127 pounds (= say) with a standard deviation of 15 pounds(= s say). The women's measurements were independent of each other.
We are to test whether the mean weight for 21-year old vegetarian women is significantly less than 132, using a significance level of 5%.
Similar to the test of hypothesis in the previous question we are to conduct a test on :
versus
Assuming the population to be Normally distributed and our samples being drawn independently,the test statistic that must be used to conduct the test on our hypothesis is called the t-Test.
t is defined as ,
where t follows a t-distribution with (n-1) degrees of freedom
i.e. t~tn-1
For our given data, the observed value of t is :
= -1.972
Now whether we should accept our null hypothesis or not is based on the camparison of our observed t value with the t value obtained from tables at 5% (=0.05) level of significance with 34 degrees of freedom (n-1=35-1).
From table, t0.05;34= 1.6909 (We use 0.05 since our H1 is one sided)
Since our |observed t| > t0.05;34 , we reject our Ho and conclude that the mean weight for 21-year old vegetarian women is significantly less than 132, using a significance level of 5%.