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Question Submission Form

All right so let here in part a the confidence interval uh is nothing but d plus minus t sub alpha by 2 times the standard deviation divided by square root n okay where d is the sum of d i by n okay and standard deviation is uh square root of the sum of d i square minus the sum of d i square divided by n whole divided by n minus 1 okay all right so here a

Is equal to 1 minus confidence level confidence level divided by 100 and t sub alpha by 2 is t table value okay t table value and confidence interval is this and d is the sum of d i by n d is the sum of d i by n which is 11.1 divided by 12 this is 0.925 okay all right and the pool standard deviation is uh sp which is uh square root of 11.1 square divided

By 12 divided by 11 so this is 0.27 okay and the confidence level is 0.92 pi plus minus t sub alpha by 2 0.156 divided by square root 12 okay so this is equal to 0.925 minus minus 3.106 times 0.078 and 0.925 plus 3.106 times 0.078 okay so this is 0.683 and 1.167 okay in part b we are given that the null hypothesis mu sub d is zero and the alternate

Hypothesis uh this is mu sub d is greater than zero okay and level of significance alpha is 0.01 and from the standard normal table the right tail t of alpha by 2 is 2.821 since our test is right tailed okay so we will reject h naught t naught is greater than 2.821 so we use the t test statistic okay uh so this is d divided by square s sorry this is s s divided

By square root n okay uh where value of s square is uh the sum of d i squared minus the sum of d i squared divided by n whole divided by n minus 1 okay so d is equal to x i minus y i divided by n this is 0.97 and we have that d is 0.97 okay and the pool variance to calculate the value of the standard deviation is square root a square so this is uh square root

Of 10.09 minus 9.7 square divided by 10 whole divided by 9 this is 0.275 okay and t naught is d divided by s divided by square root in which is eleven point one five one okay and the critical value uh uh t sub alpha is within minus one degrees of freedom is nine so the degrees of freedom is 2.821 okay and t naught modulus is 11.151 and t alpha modulus is 2.821

Okay so to make the decision hence the value of t naught is greater than t alpha okay and here we reject the null hypothesis so the p value of the right tail test is uh h sub a which is p is greater than 11.1511 okay and this is zero so hence the value of p at zero point zero one is greater than zero here we reject the null hypothesis null hypothesis okay and

Uh the null hypothesis uh uh mu sub d is equal to 0 this is the null hypothesis and we know that so here we reject the null hypothesis so we have enough evidence to support the claim that the glyph side lowers the mean hb hba1c level up people with the type 2 zebra bits okay

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Effectiveness of a medication called Glipizide in lowering the gl By WhiteBoard