Question

A university is interested in whether there's a difference between students who live on

Answer 1

It is to be noted that the determination of whether or not there is a significant difference between the two groups will be done using a t test.

What is a t test?

A t-test is a statistical test that juxtaposes two samples' means. It is used in hypothesis testing, using a null hypothesis that the variance in group means is zero and an alternative hypothesis that the difference is not zero.

What are the conditions for using this kind of confidence interval?

The conditions to use the t test are:

• The sample must be independent
• The mean of the population and variance must be unknown.
• The Box plot is attached.

What are the degrees of freedom (k) for this test using the conservative method?

The degrees of freedom (k) to be utilized for this text will be derived using the conservative method given below:

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

= [(3.0952/15) + (6.4095/15)]² / [((3.0952/15)²/14) + ((6.4095/15)²/14)]

= 24.965

Hence,

df ≈ 24 (if approximated to the floor)

What are the sample statistics for this test?

Recall the the standard deviation of the population are unequal and unknown. This thus requires that we utilize the two-sample unpooled t-test.

Here, H₀ is given as;

$t = \frac{\bar{x_{1} -\bar{x_{2}}}}{\sqrt{\frac{s_{1}^{2} }{n_{1} } + \frac{S_{2}^{2} }{n_{2}} } } \sim t_{df}$

t = [(3.33333 - 4.13333)]/√[(3.0952/15) + (6.4095/15)]

= - 0.8/√0.6337

t = - 1.005

What is the 95% confidence interval for the difference between the number of classes missed by each group of students?

The 95% confidence interval is computed using the following formula:

$(\bar x_{2} - \bar x_{1}) \pm t_\alpha_/_2_,_df \left({\sqrt{\frac{S_{1}^{2} }{n_{1} } + \frac{S_{2}^{2} }{n_{2}} } } \right)$

= - 0.8 ± t₀.₀₂₅,₂₄ (√0.6337)

= - 0.8 ± 2.064 (√0.6337)

= -2.4429, 0.8429

What is the a 90% confidence interval for the difference between the number of classes missed by each group of students?

To derive the 90% interval, we state:

$(\bar x_{2} - \bar x_{1}) \pm t_\alpha_/_2_,_df \left({\sqrt{\frac{S_{1}^{2} }{n_{1} } + \frac{S_{2}^{2} }{n_{2}} } } \right)$

= - 0.8 ± t₀.₀₅₀,₂₄ (√0.6337)

= - 0.8 ± 0.685 (√0.6337)

= -2.162, 0.562

Based on the two confidence intervals computed in parts d and e, what is the conclusion about the differences between the means of the two groups?

From the intervals computed, we must fail to reject H₀

H₀ : μ₁ = μ₂

It is clear from the above intervals computed from that the differences between the mean of both groups is significant. This is because, zero is included on the two intervals.

Learn more about t-test at;
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