This document contains the results of a group project analyzing survey data about lemonade brands. It includes: - Descriptions of a low-end, mid-tier, and premium lemonade brand. - Confidence intervals and hypothesis tests calculating the proportion of people preferring the premium brand, and whether preferences differ by gender. - Additional hypothesis tests comparing brand ratings and examining relationships between ratings, brands, gender, and other variables. - Results of t-tests and chi-squared tests finding some significant differences between brands but no relationships between other variables like gender and brand preference. - An overall conclusion that, based on the survey responses at a 95% confidence level, people rate the brands

1. Business Statistics:
Group Project
By Peter Middlemist, Zach Harris, and Mason McCoy

2. Low-End Brand
Kroger Lemonade Mix
Mid-Tier Brand
Simply Lemonade
Premium Brand
Hubert’s Original Lemonade
Products

3. What proportion of people that prefer the premium brand?
Confidence Interval
Confidence Interval: 𝑝 ± 𝑧∗ 𝑝 1− 𝑝
𝑛
0.1333 ± 1.96
0.1333 0.8667
30
= 0.1333 ± 0.1216 = (0.0117, 0.2549)
Conclusion
We are 95% confident that the true proportion of people that prefer the premium brand of
lemonade, Hubert’s, to be between 1.17% and 25.49%.
𝑝 =
4
30
= 13.33% df = n-1 = 29𝑧∗
= 1.96

4. Do more than 10% of people favor the premium brand (Hubert’s)?
Hypothesis Test for Proportions
Hypothesis
• Ho: π ≤ .10
• Ha: π > .10
• α = .05
Variables
• p= 4/30 = .1333
• π = .10
• n = 30
Test
Statistic
• Z* = (p – π)/√(π(1- π)/n)
• Z* = (.1333 – .10)/√(.1(1- .1)/30)
• Z* = .60797
Critical Value = 1.645 @ α = .05
We fail to reject the null hypothesis
that the true proportion of the
population that favors the premium
brand is less than or equal to 10% at
a 5% LOS.
C.V.Z*

5. Is the true mean rating for the low-end brand (Kroger) less than a 6?
Hypothesis Test for Mean
Hypothesis
• Ho: μ < 6
• Ha: μ ≥ 6
• α = .05
Variables
• X̄ = 6.3667
• S = 1.9561
• n = 30
• df = 29
Test
Statistic
• T* = (x̄ - μ)/(S/(√(n))
• T* = (6.3667-6)/(1.9561/(√(30))
• T* = 1.03
Critical Value = 1.699 @ α = .05
At an alpha of .05, we fail to reject
the null hypothesis that the true
mean rating for the low-end
powdered lemonade is less than 6.
Further, if we expand this test to a
10% level of significance, we must
still fail to reject the null hypothesis.
C.V.T*

6. Do males and females rate differently overall?
Two Sample t Test
Unequal Variance t Test
CV = t0.025,25 = 2.060
𝑥 𝐹 = 6.106 𝑉𝐹 = 1.919
𝑥 𝑀 = 6.166 𝑉 𝑀 = 0.444
t =
6.166−6.106
1.919
22
+
0.444
8
= 𝟎. 𝟏𝟓𝟖df = 25.379 ≈ 25
As the test statistic, at 0.158, is less extreme than 2.060, we do
not reject the null hypothesis. At a 95% confidence level, we can
conclude that there is not enough evidence to say the mean
ratings of males and females are significantly different.
After using the F distribution…
H0: µ1 = µ2
H1: µ1 ≠ µ2

7. Is there a dependency between gender and favorite brand?
Chi-Squared Test
Simply Hubert’s Kroger Total
Male 2.67 2.67 2.67 8
Female 7.33 7.33 7.33 22
Total 10 10 10 30
Simply Hubert’s Kroger Total
Male 5 1 2 8
Female 13 3 6 22
Total 18 4 8 30
Expected
Observed
𝜒2
=
𝑖=1
𝑘
𝑓𝑖 − 𝑒𝑖
2
𝑒𝑖
𝐶𝑉 = 𝜒2
.05,2 = 5.99
𝝌 𝟐
= 𝟏𝟎. 𝟒𝟑
df = 𝑟 − 1 𝑐 − 1 = 2
H0: All proportions are equal to their expected values
H1: At least one proportion is not equal to the specified value
As χ2
= 10.43, we reject the null hypothesis. There is
sufficient evidence of a relationship between one’s gender
and one’s favorite brand of lemonade.

8. Is there a dependency between whether asked to buy favorite in a store or from a lemonade stand, and
response?
Chi-Squared Test
Store Stand Total
Yes 11.5 11.5 23
No 3.5 3.5 7
Total 15 15 30
Store Stand Total
Yes 12 11 23
No 3 4 7
Total 15 15 30
Expected
Observed
𝐶𝑉 = 𝜒2
.05,1 = 3.84
H0: All proportions are equal to their expected values
H1: At least one proportion is not equal to the specified value
𝝌 𝟐
= 𝟎. 𝟏𝟖𝟔𝟑
df = 𝑟 − 1 𝑐 − 1 = 1
As χ2
= 0.1863, we fail to reject the null hypothesis. There is not
sufficient evidence of a relationship between one’s willingness to
buy from a lemonade stand vs. a traditional supermarket.

9. Is there a relationship between age (x) and overall mean lemonade rating (y)?
Linear Regression
𝑟2
= 0.014
𝑟 = 0.118
𝑦 = 3.2098 + 0.1454𝑥
H0: b1 = 0
H1: b1 ≠ 0
SE = 1.257
df = n-2 = 28
𝑡∗
=
𝑏1
𝑆𝐸
= 0.1156
𝐶𝑉 = 𝑡0.05,28 = 2.059
As 𝑡∗
= 0.1156 , we do not reject the null
hypothesis. There is not enough evidence to
suggest a relationship between age and overall
mean lemonade rating

10. Is there a relationship between sweet tooth rating (x) and overall mean ice cream rating (y)?
Linear Regression
𝑟2
= 0.035
𝑟 = 0.187
𝑦 = 9.796 − 0.349𝑥
H0: b1 = 0
H1: b1 ≠ 0
SE = 1.245
df = n-2 = 28
𝑡∗
=
𝑏1
𝑆𝐸
= −0.028
𝐶𝑉 = 𝑡0.05,28 = 2.059
As 𝑡∗
= −0.028 , we do not reject the null
hypothesis. There is not enough evidence to
suggest a relationship between age and overall
mean lemonade rating

11. Is there a difference in rating between Kroger mix and Simply Lemonade?
Two Sample t Test
Equal Variance t Test
CV = t0.05,58 = 1.671
𝑥 𝑆 = 7.433 𝑉𝑆 = 3.0126
𝑥 𝐾 = 6.367 𝑉𝐾 = 3.826
t =
7.433−4.567
3.42
1
30
+
1
30
= 𝟐. 𝟐𝟑𝟒
As the test statistic, at 2.234, is more extreme than 1.671, we
reject the null hypothesis. At a 95% confidence level, there is
enough evidence to suggest that there is a difference in the mean
ratings of Kroger mix and Simply Lemonade.
After using the F distribution…
H0: µ1 = µ2
H1: µ1 ≠ µ2
𝑃𝑉𝐸 = 3.42

12. Is there a difference in rating between Kroger mix and Hubert’s?
Two Sample t Test
CV = t0.05,58 = 1.671
𝑥 𝐻 = 4.567 𝑉𝐻 = 3.633
𝑥 𝐾 = 6.367 𝑉𝐾 = 3.826
t =
6.367−4.567
3.730
1
30
+
1
30
= 𝟑. 𝟔𝟏
As the test statistic, at 3.61, is more extreme than 1.671, we
reject the null hypothesis. At a 95% confidence level, there is
enough evidence to suggest that there is a difference in the mean
ratings of Kroger mix and Hubert’s lemonades.
Equal Variance t Test
After using the F distribution…
H0: µ1 = µ2
H1: µ1 ≠ µ2
𝑃𝑉𝐸 = 3.73

13. Is there a difference in rating between Simply Lemonade and Hubert’s?
Two Sample t Test
CV = t0.05,58 = 1.671
t =
7.433−4.567
3.32
1
30
+
1
30
= 𝟔. 𝟏𝟎
𝑥 𝑆 = 7.433
𝑥 𝐻 = 4.567
𝑉𝑆 = 3.0126
𝑉𝐻 = 3.633
As the test statistic, at 6.1, is more extreme than 1.671, we reject
the null hypothesis. At a 95% confidence level, there is enough
evidence to suggest that there is a difference in the mean ratings
of Simply Lemonade and Hubert’s lemonades.
Equal Variance t Test
After using the F distribution…
H0: µ1 = µ2
H1: µ1 ≠ µ2
𝑃𝑉𝐸 = 3.32

14. Low-End Brand
Kroger Lemonade Mix
Mid-Tier Brand
Simply Lemonade
Premium Brand
Hubert’s Original Lemonade
After running the t tests…
Based on survey responses at a 95% Confidence Level
< <

15. Is there a dependency between sweet tooth (x) and favorite brand (y)?
Chi-Squared Test
Kroger Simply Hubert’s Total
3-5 2 3 1 6
6-8 3 8 1 12
9-10 3 7 2 12
Total 8 18 4 30
Kroger Simply Hubert’s Total
3-5 1.6 3.6 0.8 6
6-8 3.2 7.2 1.6 12
9-10 3.2 7.2 1.6 12
Total 8 18 4 30
Expected
Observed
H0: All proportions are equal to their expected values
H1: At least one proportion is not equal to the specified value
𝐶𝑉 = 𝜒2
.05,4 = 9.488
𝝌 𝟐
= 𝟎. 𝟔𝟗𝟒
df = 𝑟 − 1 𝑐 − 1 = 4
As χ2
= 0.694, we fail to reject the null hypothesis.
There is not sufficient evidence of a relationship
between one’s sweet tooth rating and one’s favorite
brand of lemonade.

16. Beta Analysis: Tesla Motors Inc. (TSLA)
Date
TSLA
Close
S&P
Close TSLA % S&P %
4/1/2016 254.51 2080.73 11% 1%
3/1/2016 229.77 2059.74 20% 7%
2/1/2016 191.93 1932.23 0% 0%
1/4/2016 191.2 1940.24 -20% -5%
12/1/2015 240.01 2043.94 4% -2%
11/2/2015 230.26 2080.41 11% 0%
10/1/2015 206.93 2079.36 -17% 8%
9/1/2015 248.4 1920.03 0% -3%
8/3/2015 249.06 1972.18 -6% -6%
7/1/2015 266.15 2103.84 -1% 2%
6/1/2015 268.26 2063.11 7% -2%
5/1/2015 250.8 2107.39 11% 1%
4/1/2015 226.05 2085.51 20% 1%
3/2/2015 188.77 2067.89 -7% -2%
2/2/2015 203.34 2104.5 0% 5%
1/2/2015 203.6 1994.99 -8% -3%
12/1/2014 222.41 2058.9 -9% 0%
11/3/2014 244.52 2067.56 1% 2%
10/1/2014 241.7 2018.05 0% 2%
9/2/2014 242.68 1972.29 -10% -2%
8/1/2014 269.7 2003.37 21% 4%
7/1/2014 223.3 1930.67 -7% -2%
6/2/2014 240.06 1960.23 16% 2%
5/1/2014 207.77 1923.57 0% 2%
4/1/2014 207.89 1883.95 0% 1%
3/3/2014 208.45 1872.34 -15% 1%
2/3/2014 244.81 1859.45 35% 4%
1/2/2014 181.41 1782.59 21% -4%
Selected Data
0
500
1000
1500
2000
2500
Nov-10 Apr-12 Aug-13 Dec-14 May-16 Sep-17
Price Movement (4/1/11-4/1/16)
TSLA Close S&P Close
Formula = COVARIANCE.S(% Change TSLA, % Change ^GSPC)/VAR.S(% Change ^GSPC)
-10%
-5%
0%
5%
10%
15%
-50% 0% 50% 100%
^GSPCReturn
TSLA Return
Scatterplot of ^GSPC & TSLA
Beta = 1.331135
Tesla (TSLA) vs. S&P 500 (^GSPC)
April 1, 2011 through April 1, 2016
Monthly, 60 Periods
Therefore, we can conclude that for every 1% change in the price of the S&P 500 (^GSPC), on
average there will be a 1.33% increase in the price of one share of Tesla (TSLA).

17. Thank You!