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Question

Test the given claim below. Use CI or p-value when indicated, Find the null, alternative hypothesis, test statistic, critical value(s) or the p-value (or the range of p-values) as appropriate, and state the final conclusion that address the original claim

1) A cereal company claims that the mean weight  of the cereal in its packets is 14oz. The weights (in ounces) of the cereal in a random sample of 7 of its cereal packets are listed below.

14.6 13.8 14.1 13.7 14.0 14.4 13.6

Test the claim at the 0.01 signifcance level.

2) A public bus company offical claims that the mean waiting time for bus number 14 during peak hours is less than 10 minutes. Karen took bus number 14 during peak hours on 18 different occasions. Her mean waiting time was 7.6 minutes with a standard deviation of 2.3 minutes. At the 0.01 signficance level, test the claim that the mean waiting time is less than 10 minutes. Use the p-value method of testing hypotheses.

Answer 1

Step 1:

(1)

A cereal company claims that the mean weight  of the cereal in its packets is 14oz.

Null hypothesis :

One of these statements must become the null hypothesis , and the other should be the alternate hypothesis.

The null hypothesis contains equality.

So for the above, the null hypothesis H₀ : x = 14.

Step 2:

Alternate hypothesis :

The statement that does not contain equality is the alternative hypothesis.

A cereal company claims mean weight which is not equal to 14oz.

So for the above, the alternate hypothesis H_a : x ≠ 14.

Step 3:

Find the test statics.

The sample size of the cereal packets is 7.

Sample contains cereal packets of weights : 14.6, 13.8, 14.1, 13.7, 14.0, 14.4, 13.6.

Calculate mean and standard deviation of the sample.

Follow these steps to evaluate Mean and standard deviation.

1.First enter the sample values.

[STAT --> 1 --> L1 ]

Enter the values in the L1 column.

2.Select 1-variable stats

[STAT --> Right navigation key --> ENTER]

Var list L1.

3.Now press Enter in calculator to view answer.

Mean = 14.02857

Standard deviation = 0.36839.

Test statics : .

 

Test statics : .

Step 4:

Find the critical values.

Significance value is 0.01.

.

The sample size of the cereal packets is 7.

Degree of freedom .

Calculate the critical values using calculator.

Follow these steps to evaluate critical values.

1.Select invT()

[2nd --> VARS --> 4 ]

2.Enter the values of and df.

area : 0.01

df : 6

3.Now press Enter in calculator to view answer.

invT(0.01, 6)

=-3.14266.

Critical value is -3.14266.

 

Comments

Step 5:

Find the p-value.

Follow these steps to evaluate p-value.

1.Select tcdf()

[2nd --> VARS --> 6 ]

2.Enter the values of t and df.

Lower : 0.20518

Upper : 1000

df : 6

For two-tailed (non-directional), the upper value is considered as 1000.

3.Now press Enter in calculator to view answer.

tcdf(0.20518, 1000, 6)

=0.4221

p-Value of the hypotheses is 0.4221

Step 6:

Conclusion:

Since the value of test statics is less than critical values, it fails to reject H₀.

The test results support the company claim.

Solution:

The null hypothesis H₀ : x = 14.

The alternate hypothesis H_a : x ≠ 14.

Test statics : .

Critical values is -3.14266.

p-value of the hypothesis is 0.34434.

The test results support the company claim.

Answer 2

Step 1:

(2)

A public bus company official claims that the mean waiting time for bus number 14 is less than 10 minutes.

Null hypothesis :

One of these statements must become the null hypothesis , and the other should be the alternate hypothesis.

The null hypothesis contains equality.

So for the above, the null hypothesis H₀ : x = 10.

Step 2:

Alternate hypothesis :

The statement that does not contain equality is the alternative hypothesis.

Mean waiting time for bus number 14 is not equal 10 minutes.

So for the above, the alternate hypothesis H_a : x ≠ 10.

Step 3:

Karen took bus number 14 during peak hours on 18 different occasions.

So the number of samples is 18.

Mean waiting time was 7.6 minutes.

Standard deviation is 2.3 minutes.

Test statics : .

Test statics : .

Comments

Step 4:

Find the critical values.

Significance value is 0.01.

.

The number of samples is 18.

Degree of freedom .

Calculate the critical values using calculator.

Follow these steps to evaluate critical values.

1.Select invT()

[2nd --> VARS --> 4 ]

2.Enter the values of and df.

area : 0.01

df : 17

3.Now press Enter in calculator to view answer.

invT(0.01, 17)

Step 5:

Find the p-value.

Follow these steps to evaluate p-value.

1.Select tcdf()

[2nd --> VARS --> 6 ]

2.Enter the values of t and df.

Lower : 4.427

Upper : 1000

df : 17

For two-tailed (non-directional), the upper value is considered as 1000.

3.Now press Enter in calculator to view answer.

tcdf(4.427, 1000, 17)

=0.000184.

.

p-Value of the hypothesis is 0.000184.

Step 6:

Conclusion:

Since p -value = , we should reject H₀ .

We conclude that the mean is less than 10 minutes.

Solution:

The null hypothesis H₀ : x = 10.

The alternate hypothesis H_a : x ≠ 10.

Test statics : .

Critical values is .

p-value of the hypothesis is 0.000184.

The mean is less than 10 minutes.