Question

A new baker is trying to decide if he has an appropriate price set for his 3 tier wedding cakes which he sells for $84.3. He is particullarly interested in seeing if his wedding cakes do not sell for the average price. He searches online and finds 18 of the competitors in his area that sell 3 tier wedding cakes with a mean price of $98.72 with a standard deviation of $6.65. Help the new baker by testing his his claim with a 0.01 level of significance.
The correct hypotheses would be:

• H0:μ≤$84.3H0:μ≤$84.3
  HA:μ>$84.3HA:μ>$84.3 (claim)
• H0:μ≥$84.3H0:μ≥$84.3
  HA:μ<$84.3HA:μ<$84.3 (claim)
• H0:μ=$84.3H0:μ=$84.3
  HA:μ≠$84.3HA:μ≠$84.3 (claim) Since the level of significance is 0.01 the critical value is 2.898 and -2.898

The test statistic is: (round to 3 places)
The p-value is: (round to 3 places)

A recent study showed that the average number of sticks of gum a person chews in a week is 15. A college student believes that the guys in his dormitory chew more gum in a week. He conducts a study and samples 23 of the guys in his dorm and finds that on average they chew 13 sticks of gum in a week with a standard deviation of 2.9. Test the college student's claim at αα=0.10.
The correct hypotheses would be:

• H0:μ≤15H0:μ≤15
  HA:μ>15HA:μ>15 (claim)
• H0:μ≥15H0:μ≥15
  HA:μ<15HA:μ<15 (claim)
• H0:μ=15H0:μ=15
  HA:μ≠15HA:μ≠15 (claim)

Since the level of significance is 0.10 the critical value is 1.321
The test statistic is: (round to 3 places)
The p-value is: (round to 3 places)

A recent publication states that the average closing cost for purchasing a new home is $8929. A real estate agent believes the average closing cost is different than $8929. She selects 22 new home purchases and finds that the average closing costs are $7298 with a standard deviation of $184. Help her decide if she is correct by testing her claim at αα=0.05.
The correct hypotheses would be:

• H0:μ≤$8929H0:μ≤$8929
  HA:μ>$8929HA:μ>$8929 (claim)
• H0:μ≥$8929H0:μ≥$8929
  HA:μ<$8929HA:μ<$8929 (claim)
• H0:μ=$8929H0:μ=$8929
  HA:μ≠$8929HA:μ≠$8929 (claim)

Since the level of significance is 0.05 the critical value is 2.08 and -2.08
The test statistic is: (round to 3 places)
The p-value is: (round to 3 places)

A new baker is trying to decide if he has an appropriate price set for his 3 tier wedding cakes which he sells for $88.93. He is particullarly interested in seeing if his wedding cakes do not sell for the average price. He searches online and finds that out of 39 of the competitors in his area they sell their 3 tier wedding cakes for $90.19. From a previous study he knows the standard deviation is $8.97. Help the new baker by testing this with a 0.01 level of significance.
The correct hypotheses would be:

• H0:μ≤$88.93H0:μ≤$88.93
  HA:μ>$88.93HA:μ>$88.93 (claim)
• H0:μ≥$88.93H0:μ≥$88.93
  HA:μ<$88.93HA:μ<$88.93 (claim)
• H0:μ=$88.93H0:μ=$88.93
  HA:μ≠$88.93HA:μ≠$88.93 (claim)

Since the level of significance is 0.01 the critical value is 2.576 and -2.576
The test statistic is: (round to 3 places)
The p-value is: (round to 3 places)

A recent publication states that the average closing cost for purchasing a new home is $8996. A real estate agent believes that the average closing cost is more than $8996. She selects 30 new home purchases and finds that the average closing costs are $7199. The population standard deviation of $428. Help her decide if she is correct by testing her claim at αα=0.05.
The correct hypotheses would be:

• H0:μ≤$8996H0:μ≤$8996
  HA:μ>$8996HA:μ>$8996 (claim)
• H0:μ≥$8996H0:μ≥$8996
  HA:μ<$8996HA:μ<$8996 (claim)
• H0:μ=$8996H0:μ=$8996
  HA:μ≠$8996HA:μ≠$8996 (claim)

Since the level of significance is 0.05 the critical value is 1.645
The test statistic is: (round to 3 places)
The p-value is: (round to 3 places)

Answer 1

Ans 1 ) using excel addin>phstat<one sample test

we have

t Test for Hypothesis of the Mean
Data
Null Hypothesis                m=	84.3
Level of Significance	0.01
Sample Size	18
Sample Mean	98.72
Sample Standard Deviation	6.65
Intermediate Calculations
Standard Error of the Mean	1.5674
Degrees of Freedom	17
t Test Statistic	9.1998
Two-Tail Test
Lower Critical Value	-2.8982
Upper Critical Value	2.8982
p-Value	0.0000
Reject the null hypothesis

The correct hypotheses would be:
H0:μ=$84.3

HA:μ≠$84.3 (claim) Since the level of significance is 0.01 the critical value is 2.898 and -2.898

The test statistic is: 9.200
The p-value is:0.000

Ans 2 ) using excel>addin>phstat>one sample test

we have

t Test for Hypothesis of the Mean
Data
Null Hypothesis                m=	15
Level of Significance	0.1
Sample Size	23
Sample Mean	13
Sample Standard Deviation	2.9
Intermediate Calculations
Standard Error of the Mean	0.6047
Degrees of Freedom	22
t Test Statistic	-3.3075
Upper-Tail Test
Upper Critical Value	1.3212
p-Value	0.9984
Do not reject the null hypothesis

The correct hypotheses would be:

• H0:μ≤15
  HA:μ>15 (claim)

Since the level of significance is 0.10 the critical value is 1.321
The test statistic is: -3.307
The p-value is: 0.998

Ans 3 ) using excel>addin>phstat>one sample

we have

t Test for Hypothesis of the Mean
Data
Null Hypothesis                m=	8929
Level of Significance	0.05
Sample Size	22
Sample Mean	7298
Sample Standard Deviation	184
Intermediate Calculations
Standard Error of the Mean	39.2289
Degrees of Freedom	21
t Test Statistic	-41.5765
Two-Tail Test
Lower Critical Value	-2.0796
Upper Critical Value	2.0796
p-Value	0.0000
Reject the null hypothesis

H0:μ=$8929H0:μ=$8929
HA:μ≠$8929HA:μ≠$8929 (claim)

Since the level of significance is 0.05 the critical value is 2.08 and -2.08
The test statistic is:-41.576
The p-value is:0.000

Ans 4 )using excel>addin>phstat>one sample test

we have

t Test for Hypothesis of the Mean
Data
Null Hypothesis                m=	88.93
Level of Significance	0.01
Sample Size	39
Sample Mean	90.19
Sample Standard Deviation	8.97
Intermediate Calculations
Standard Error of the Mean	1.4363
Degrees of Freedom	38
t Test Statistic	0.8772
Two-Tail Test
Lower Critical Value	-2.7116
Upper Critical Value	2.7116
p-Value	0.3859
Do not reject the null hypothesis

• H0:μ=$88.93H0:μ=$88.93
  HA:μ≠$88.93HA:μ≠$88.93 (claim)

Since the level of significance is 0.01 the critical value is 2.576 and -2.576
The test statistic is: 0.877
The p-value is: 0.386

Ans 5) using excel>Addin>phstat>one sample test

we have

• Z Test of Hypothesis for the Mean
  Data
  Null Hypothesis                       m=	8996
  Level of Significance	0.05
  Population Standard Deviation	428
  Sample Size	30
  Sample Mean	7199
  Intermediate Calculations
  Standard Error of the Mean	78.1418
  Z Test Statistic	-22.9967
  Upper-Tail Test
  Upper Critical Value	1.6449
  p-Value	1.0000
  Do not reject the null hypothesis

  Since the level of significance is 0.05 the critical value is 1.645
• H0:μ≤$8996H0:μ≤$8996
  HA:μ>$8996HA:μ>$8996 (claim)

The test statistic is:-22.997
The p-value is: 1.0000