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Gekata [30.6K]
4 years ago
8

A recent study reported that high school students spend an average of 94 minutes per day texting. Jenna claims that the average

for the students at her large high school is greater than 94 minutes. She will conduct a study to investigate this claim.
(b) Based on a sample of 32 students, Jenna calculated a sample mean of 96.5 minutes and a sample standard deviation of 6.3 minutes. Assume all conditions for inference are met. At the significance level of α=0.05, do the data provide convincing statistical evidence to support Jenna’s claim? Complete an appropriate inference procedure to support your answer.
Mathematics
1 answer:
larisa [96]4 years ago
3 0

Answer:

We have sufficient evidence to support the claim that the average for the students at Jenna's large high school is greater than 94 minutes.

Step-by-step explanation:

Jenna claims that the average time of texting at her larger high school is greater than 94 minutes per day.

From here we can see that we have to perform a hypothesis test about a sample mean. The null and alternate hypothesis will be:

Null Hypothesis: \mu \leq  94

Alternate Hypothesis: \mu > 94

Jenna collected data from a sample of 32 students. So, sample size will be:

Sample Size = n = 32

Sample Mean = x = 96.5

Sample Standard Deviation = s = 6.3

We have to perform a hypothesis test, to test Jenna's claim. Since, the value of Population Standard Deviation is unknown and the value of Sample Standard Deviation is known, we will use One Sample t-test in this case.

The formula to calculate the test statistic is:

t=\frac{x-\mu}{\frac{s}{\sqrt{n}}}

Using the values, we get:

t=\frac{96.5-94}{\frac{6.3}{\sqrt{32} } }=2.245

The degrees of freedom will be:

df = n - 1 = 32 - 1 = 31

We have to convert the t-score 2.245 with 31 degrees of freedom to its equivalent p-value. From t-table this value comes out to be:

p-value = 0.0160

The significance level is:

\alpha =0.05

Since, the p-value is lesser than the level of significance, we reject the Null Hypothesis.

Conclusion:

We have sufficient evidence to support the claim that the average for the students at Jenna's large high school is greater than 94 minutes.

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90% confidence interval is ( -149.114, -62.666   )

Step-by-step explanation:

Given the data in the question;

Sample 1                                Sample 2

x"₁ = 259.23                            x"₂ = 365.12

s₁  = 34.713                              s₂ = 48.297

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With 90% confidence interval for μ₁ - μ₂ { using equal variance assumption }

significance level ∝ = 1 - 90% = 1 - 0.90 = 0.1

Since we are to assume that variance are equal and they are know, we will use pooled variance;

Degree of freedom DF = n₁ + n₂ - 2 = 5 + 10 - 2 = 13

Now, pooled estimate of variance will be;

S_p^2 = [ ( n₁ - 1 )s₁² + ( n₂ - 1)s₂² ] / [ ( n₁ - 1 ) + ( n₂ - 1 ) ]

we substitute

S_p^2 = [ ( 5 - 1 )(34.713)² + ( 10 - 1)(48.297)² ] / [ ( 5 - 1 ) + ( 10 - 1 ) ]

S_p^2 = [ ( 4 × 1204.9923) + ( 9 × 2332.6 ) ] / [  4 + 9 ]

S_p^2 = [ 4819.9692 + 20993.4 ] / [  13 ]

S_p^2 = 25813.3692 / 13

S_p^2 = 1985.64378

Now the Standard Error will be;

S_{x1-x2 = √[ ( S_p^2 / n₁ ) + ( S_p^2 / n₂ ) ]

we substitute

S_{x1-x2 = √[ ( 1985.64378 / 5 ) + ( 1985.64378 / 10 ) ]

S_{x1-x2 = √[ 397.128756 + 198.564378 ]

S_{x1-x2 = √595.693134

S_{x1-x2 = 24.4068

Critical Value = t_{\frac{\alpha }{2}, df = t_{0.05, df=13 = 1.771  { t-table }

So,

Margin of Error E =  t_{\frac{\alpha }{2}, df × [ ( S_p^2 / n₁ ) + ( S_p^2 / n₂ ) ]

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Margin of Error E = 1.771 × 24.4068

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Lower Limit = x₁ - x₂ - E = -105.89 - 43.224 = -149.114

Upper Limit = x₁ - x₂ - E = -105.89 + 43.224 = -62.666

Therefore, 90% confidence interval is ( -149.114, -62.666   )

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