Question

A simple random sample of 28 filtered​ 100-mm cigarettes is obtained from a normally distributed​ population, and the tar content of each cigarette is measured. The sample has a standard deviation of 0.22 mg. Use a 0.05 significance level to test the claim that the tar content of filtered​ 100-mm cigarettes has a standard deviation different from 0.35 ​mg, which is the standard deviation for unfiltered​ king-size cigarettes. Complete parts​ (a) through​ (d) below. a. What are the null and alternative​ hypotheses? A. H0​: sigmaequals0.35 mg Upper H 1 : sigma not equals 0.35 mg Your answer is correct.B. H0​: sigmanot equals0.35 mg Upper H 1 : sigma equals 0.35 mg C. H0​: sigmaequals0.35 mg Upper H 1 : sigma less than 0.35 mg D. H0​: sigmagreater than0.35 mg Upper H 1 : sigma less than or equals 0.35 mg b. Find the test statistic. chi squaredequals 10.668 ​(Round to three decimal places as​ needed.) c. Find the​ P-value of the test statistic. The​ P-value of the test statistic is 0.002. ​(Round to three decimal places as​ needed.)

(PLEASE HELP ME FIND P-VALUE AND HOW TO PUT INTO EXCEL OR STATCRUNCH. TO FIGURE OUT THE REST OF THE HOMEWORK. THANK YOU)

Answer 1

a)

Ho :   σ =   0.35
Ha :   σ ╪   0.35

b)

Level of Significance ,    α =    0.05
sample Std dev ,    s =    0.22
Sample Size ,   n =    28
      
Chi-Square Statistic,   X² = (n-1)s²/σ² =    10.668

c)

degree of freedom,   DF=n-1 =    27

p-Value   =   0.002
[excel function: "=CHISQ.DIST(10.668,27,TRUE)"

since, p value <α=0.05, Reject the null hypothesis