STAT friday homework

A. Test of Significance
State: At the alpha .05 level, is there sufficient evidence to conclude that the average calories content of 1-ounce chocolate chip cookies produced by this brand is greater than 110 calories?
Plan: the appropriate test is one sample t test of significance for the means.
Parameter: mean calories for 1-ounce chocolate chip cookies of a certain brand of 1-ounce chocolate chip cookies that claims that the cookies contain 110 calories.
Population of interest in the context of the problem: all 1-ounce chocolate chip cookies of a certain brand of 1-ounce chocolate chip cookies that claims that the cookies contain 110 calories.
Null hypotheses: Mu=110 calories
Alternative hypotheses: Mu> 110 calories
State level of significance: alpha equals .05
Solve
plot data: with a stem leaf plot graph
stem leaf
10 0, 0
11 0
12 0, 5, 5
13 5
14 0, 5
15 0, 0, 5
16 0, 0
17
18 5
Shape: the five number min: 100; max: 185; x bar: 137.3333; 1Q:115 ;3Q: 155.625
It is approximately normal
List and check conditions of inference: the two conditions are randomization, and normality of the population both of these conditions are not meant. The randomization is not false because the sample was a srs, and we learned that from the provided story problem information. The normality of the population is not true because from the plotted data it can be seen that the population is relatively normal.
Calculate the value(s) of the sample statistic(s): used Stat crunch
N=15
Ho Mu=110
Ha Mu does not equal 110
X bar=137.3333
Std. Err. = 6.22718
T-Stat= 4.3893595
Std. Dev. = 24.117767
P-value =.0006
[(X bar=137.3333)- (110)]/[(s)/(3.872983)]=t=4.38935
100+125+150+160+185+125+155+145+160+100+150+140+135+120+110=2,060
2,060/15= 137.3333= x bar
Conclude:
The set alpha was equal to .05 and the p-value of .0006 was lower than the set alpha, because the p-value was significantly lower than the set alpha I chose to reject the Ho. There is sufficient evidence to conclude that the average calories content of 1-ounce chocolate chip cookies produced by this brand is greater than 110 calories.

Confidence Interval Estimation
State: compute a 95% confidence interval to estimate the mean amount of calories for all 1-ounce chocolate chip cookies of this brand.
Plan:
State the name of the appropriate estimation procedure: this is a one sample t test of 95% confidence to of the means.
Describe the parameter: Parameter: mean calories for 1-ounce chocolate chip cookies of a certain brand of 1-ounce chocolate chip cookies that claims that the cookies contain 110 calories.
Population of interest in the context of the problem: all 1-ounce chocolate chip cookies of a certain brand of 1-ounce chocolate chip cookies that claims that the cookies contain 110 calories.
State Confidence Level: the confidence level= 95% (given in problem)
Solve:
Already plotted data in part 1
Mean: 137.3333
Min: 100
Max: 185
Q1:120
Q3:155
Std. Dev. = 24.117767
Std. Err. = 6.2271
Confidence intervals: (123.977356, 150.6893)
Conclude: we believe at a confidence is 95% the true mu of a particular brand of 1-ounce chocolate chip cookies that claims a 110 calories per cookie has a confidence interval at (123.9777, 150.6893 ).