A well known pharmaceutical manufacturer is manufacturing a newly developed vaccine, and is concerned about variability in the immune response of the recipients. They claim the variance in the immune response is 1.9. It is known that the immune responses follow a normal distribution. A random sample of 30 individuals were given the vaccine and their immune response recorded. It was found that they had a sample variance of 0.9. At the 5% significance level, test the hypothesis that the population variance of the immune response is equal to 1.9 against the two-sided alternative, that it is not. a. State the null and alternative hypothesis. b. Perform the hypothesis test. (You may use any of the methods we discussed that are appropriate for the problem.) c. State your decision. d. State your conclusion in full sentences.

a

The  null hypothesis is

The alternative hypothesis is    

b

c

The decision rule is  

Reject the null hypothesis

d

The conclusion is  

There is no sufficient evidence to conclude that the variance of the immune response is equal to 1.9.

Step-by-step explanation:

From the question we are are told that

   The variance is

    The sample size is  n =  30  

    The  sample variance  is  

    The level of significance is  

The  null hypothesis is

The alternative hypothesis is    

Generally the test statistics  is mathematically represented as

=>      

=>      

Generally  from the degree of freedom is mathematically represented as

=>      

=>      

Generally from the chi distribution table the critical value of   at a degree of freedom  of   is  

=>    

Gnerally from the values obtained we see that

          hence

The decision rule is  

Reject the null hypothesis

The conclusion is  

There is no sufficient evidence to conclude that the variance of the immune response is equal to 1.9.

arithmetic sequence

step-by-step explanation:

given that a sequence has terms as

19, 15, 11,

this is represented on a graph

the pattern indicates that the sequence is an arithmetic sequence with i term

a=19 and common difference d =-4

any general term

thus the sequence is reduced by 4 for each successive term starting from 1 and nth term is

24-5n.

first negative term would start after 5th term

5th term =24-25< 0

from 5th term onwards the sequence would be negative.

what is the graph

step-by-step explanation: