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Students in a statistics class are conducting a survey to estimate the mean number of units students at their college are enrolled in. The students collect a random sample of 49 students. The mean of the sample is 12.2 units. The standard deviation is 1.6 units. What is the 95% confidence interval for the number of units students in their college are enrolled in? Assume that the distribution of individual student enrollment units at this college is approximately normal. Group of answer choices (10.2 , 14.2) (12.0 , 12.4) (9.0 , 15.4) (11.7 , 12.7) No confidence interval can be constructed since the conditions for a T-interval are not met.

Respuesta :

Answer:

[tex]12.2-2.01\frac{1.6}{\sqrt{49}}=11.7[/tex]    

[tex]12.2+2.01\frac{1.6}{\sqrt{49}}=12.7[/tex]    

And the best option would be:

(11.7 , 12.7)

Step-by-step explanation:

Information given

[tex]\bar X=12.2[/tex] represent the sample mean for the number of units

[tex]\mu[/tex] population mean (variable of interest)

s=1.6 represent the sample standard deviation

n=49 represent the sample size  

Solution to the problem

The confidence interval for the true population mean is given by:

[tex]\bar X \pm t_{\alpha/2}\frac{s}{\sqrt{n}}[/tex]   (1)

The degrees of freedom are given by:

[tex]df=n-1=49-1=48[/tex]

The Confidence required is 0.95 or 95%, the value of [tex]\alpha=0.05[/tex] and [tex]\alpha/2 =0.025[/tex], and the critical value for this case would be [tex]t_{\alpha/2}=2.01[/tex]

And the confidence interval is:

[tex]12.2-2.01\frac{1.6}{\sqrt{49}}=11.7[/tex]    

[tex]12.2+2.01\frac{1.6}{\sqrt{49}}=12.7[/tex]    

And the best option would be:

(11.7 , 12.7)

Using the t-distribution, it is found that the 95% confidence interval for the number of units students in their college are enrolled in is (11.7, 12.7).

In this question, we are given the standard deviation for the sample, thus, the t-distribution is used to build the confidence interval.

The information for the sample are as follows:

  • 49 students, thus [tex]n = 49[/tex]
  • Mean of 12.2, thus [tex]\overline{x} = 12.2[/tex].
  • Standard deviation of 1.6, thus [tex]\sigma = 1.6[/tex].

The confidence interval is:

[tex]\overline{x} \pm t\frac{s}{\sqrt{n}}[/tex]

First, we find the number of degrees of freedom, which is one less than the sample size, thus df = 48.

Then, we find the critical value for a 95% confidence interval, with 48 df, which is t = 2.0106. Then:

[tex]\overline{x} - t\frac{s}{\sqrt{n}} = 12.2 - 2.0106\frac{1.6}{\sqrt{49}} = 11.7[/tex]

[tex]\overline{x} + t\frac{s}{\sqrt{n}} = 12.2 + 2.0106\frac{1.6}{\sqrt{49}} = 12.7[/tex]

The confidence interval is (11.7, 12.7).

A similar problem is given at https://brainly.com/question/13769784

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