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A swimming school claimed that the average seven-year-old would be able to swim across an Olympic-sized pool in less than 120 seconds after taking lessons from their instructors. To test this claim, a consumer psychologist arranged for eight randomly selected seven-year-old children to take lessons at the school and recorded how long it took each child to swim across a pool at the end of the lessons.

The times (in seconds) were 60, 120, 110, 80, 70, 90, 100, and 130.

What conclusion would the psychologist draw following a t test for a single sample using 120 seconds as the "known" population mean and the .05 significance level? 13 points) across an

a. Sample mean:
b. Sample sum of squares:
c. Degrees of freedom:
d. The estimated population variance:

Respuesta :

Answer:

Step-by-step explanation:

The mean of the set of data given is

Mean = (60 + 120 + 110 + 80 + 70 + 90 + 100 + 130)/8 = 95

Standard deviation = √(summation(x - mean)/n

n = 8

Summation(x - mean) = (60 - 95)^2 + (120 - 95)^2 + (110 - 95)^2 + (80 - 95)^2 + (70 - 95)^2 + (90 - 95)^2 + (100 - 95)^2 + (130 - 95)^2 = 4200

Standard deviation = √(4200/8) = 22.91

We would set up the hypothesis test. This is a test of a single population mean since we are dealing with mean

For the null hypothesis,

µ ≤ 120

For the alternative hypothesis,

µ > 120

This is a right tailed test.

Since the number of samples is small and no population standard deviation is given, the distribution is a student's t.

Since n = 8,

Degrees of freedom, df = n - 1 = 8 - 1 = 7

t = (x - µ)/(s/√n)

Where

x = sample mean = 95

µ = population mean = 120

s = samples standard deviation = 22.91

t = (95 - 120)/(22.91/√8) = - 3.09

We would determine the p value using the t test calculator. It becomes

p = 0.009

Since alpha, 0.05 > than the p value, 0.009, then we would reject the null hypothesis. Therefore, At a 5% level of significance, the sample data showed significant evidence that the average seven-year-old would be able to swim across an Olympic-sized pool in more than 120 seconds after taking lessons from their instructors.

fichoh

Using the one - sample t test, we can conclude that average 7year old will swim in less than 120 hours

Given the data :

  • 60, 120, 110, 80, 70, 90, 100, and 130.
  • Sample size, n = 8

Using calculator :

  • Mean = ΣX/n = 760/8 = 95

  • Standard deviation = 24.49

The degree of freedom :

  • df = n - 1
  • df = 8 - 1 = 7

The hypothesis :

[tex]H_{0} : μ > 120 [/tex]

[tex]H_{1} : μ ≤ 120 [/tex]

From the equality sign in the alternative hypothesis, we have a left-tailed test

Test statistic :

[tex] \frac{x - μ}{\frac{s}{\sqrt{n}}} [/tex]

Inputting the values :

[tex] \frac{95 - 120}{\frac{24.49}{\sqrt{8}}} [/tex]

[tex] \frac{-25}{8.658} = - 2.887 [/tex]

Calculating the P-value :

  • α = 0.05 ; df = 7 ;

Pvalue = 0.012

Decison Region :

  • Reject Null if Pvalue is < α

Since 0.012 < 0.05 ; we reject [tex]H_{0} [/tex] and conclude that average 7year old will swim in less than 120 hours.

Learn more : https://brainly.com/question/16314496

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