Can someone check for me or help me get the correct answer i

Can someone check for me or help me get the correct answer idk where i go worng

A soda manufacturer must control the amount of soda in each bottle. Each bottle is supposed to contain 255mL of soda. On a given day. a random sample of 23 bottles had mean x = 256.3mL and standard deviation sx = 5.7 mL. What is the true mean soda volume mu of the soda at: a) a 95% confidence interval? b) a 90% confidence interval? c) What is the p-value of our measurements under the null hypothesis that says the average volume of a bottle is 2SSmL? (To answer this question. you will need to use Table D backwards. When doing so choose the significance level in the Table that is closest to your t-score) 2. Are people in New York taller than people in Indiana? Suppose we conducted an experiment to find out. We randomly sample 102 New Yorkers and compute a mean height of x1 = 1.3m, and a standard deviation of s1 = 0.3,n In addition we randomly sample 84 people from Indiana and compute a mean height of x2 = 1.1m, and a standard deviation of s2 = 0.4m. a) What is the p-value of our dataset under the null hypothesis that says that it is not the case that New Yorkers are taller than people in Indiana. i.e. the average difference in heights mu = 0? Use the largest conservative estimate for the degrees of freedom given in Table D. while choosing the significance level in the Table closest to your t-score Using your exact t-score from part a). can we conclude that people from New York are taller than people from Indiana at: b) The alpha = 0.05 significance level? c) The alpha = 0.1 significance level?

Solution

Q1.
1.
CI = x ± t a/2 * (sd/ Sqrt(n))
Where,
x = Mean
sd = Standard Deviation
a = 1 - (Confidence Level/100)
ta/2 = t-table value
CI = Confidence Interval
Mean(x)=256.3
Standard deviation( sd )=5.7
Sample Size(n)=23
Confidence Interval = [ 256.3 ± t a/2 ( 5.7/ Sqrt ( 23) ) ]
= [ 256.3 - 2.074 * (1.189) , 256.3 + 2.074 * (1.189) ]
= [ 253.835,258.765 ]

2.
Confidence Interval = [ 256.3 ± t a/2 ( 5.7/ Sqrt ( 23) ) ]
= [ 256.3 - 1.72 * (1.189) , 256.3 + 1.72 * (1.189) ]
= [ 254.25,258.34 ]

3.
Set Up Hypothesis
Null, H0: U=255
Alternate, H1: U>255
Test Statistic
Population Mean(U)=255
Sample X(Mean)=256.3
Standard Deviation(S.D)=5.7
Number (n)=23
we use Test Statistic (t) = x-U/(s.d/Sqrt(n))
to =256.3-255/(5.7/Sqrt(22))
to =1.094
P-Value :Right Tail - Ha : ( P > 1.0938 ) = 0.14294 ~ 0.15