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A new prep class was designed to improve SAT test scores. Five students were selected at random. Their scores on two practice exams were recorded, one before the class and one after. The data recorded in [link] . Are the scores, on average, higher after the class? Test at a 5% level.

SAT Scores Student 1 Student 2 Student 3 Student 4
Score before class 1840 1960 1920 2150
Score after class 1920 2160 2200 2100

The p -value is 0.0874, so we decline to reject the null hypothesis. The data do not support that the class improves SAT scores significantly.

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Seven eighth graders at Kennedy Middle School measured how far they could push the shot-put with their dominant (writing) hand and their weaker (non-writing) hand. They thought that they could push equal distances with either hand. The data were collected and recorded in [link] .

Distance (in feet) using Student 1 Student 2 Student 3 Student 4 Student 5 Student 6 Student 7
Dominant Hand 30 26 34 17 19 26 20
Weaker Hand 28 14 27 18 17 26 16

Conduct a hypothesis test to determine whether the mean difference in distances between the children’s dominant versus weaker hands is significant.

Record the differences data. Calculate the differences by subtracting the distances with the weaker hand from the distances with the dominant hand. The data for the differences are: {2, 12, 7, –1, 2, 0, 4}. The differences have a normal distribution.

Using the differences data, calculate the sample mean and the sample standard deviation. x ¯ d = 3.71, s d = 4.5.

Random variable: X ¯ d = mean difference in the distances between the hands.

Distribution for the hypothesis test: t 6

H 0 : μ d = 0  H a : μ d ≠ 0

Graph:

This is a normal distribution curve with mean equal to zero. Both the right and left tails of the curve are shaded. Each tail represents 1/2(p-value) = 0.0358.

Calculate the p -value: The p -value is 0.0716 (using the data directly).

(test statistic = 2.18. p -value = 0.0719 using ( x ¯ d = 3.71 ,   s d = 4.5. )

Decision: Assume α = 0.05. Since α < p -value, Do not reject H 0 .

Conclusion: At the 5% level of significance, from the sample data, there is not sufficient evidence to conclude that there is a difference in the children’s weaker and dominant hands to push the shot-put.

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Try-it

Five ball players think they can throw the same distance with their dominant hand (throwing) and off-hand (catching hand). The data were collected and recorded in [link] . Conduct a hypothesis test to determine whether the mean difference in distances between the dominant and off-hand is significant. Test at the 5% level.

Player 1 Player 2 Player 3 Player 4 Player 5
Dominant Hand 120 111 135 140 125
Off-hand 105 109 98 111 99

The p -level is 0.0230, so we can reject the null hypothesis. The data show that the players do not throw the same distance with their off-hands as they do with their dominant hands.

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Chapter review

A hypothesis test for matched or paired samples (t-test) has these characteristics:

  • Test the differences by subtracting one measurement from the other measurement
  • Random Variable: x ¯ d = mean of the differences
  • Distribution: Student’s-t distribution with n – 1 degrees of freedom
  • If the number of differences is small (less than 30), the differences must follow a normal distribution.
  • Two samples are drawn from the same set of objects.
  • Samples are dependent.

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Assimilatory nitrate reduction is a process that occurs in some microorganisms, such as bacteria and archaea, in which nitrate (NO3-) is reduced to nitrite (NO2-), and then further reduced to ammonia (NH3).
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They are friends to host only when Host immune system is strong and become enemies when the host immune system is weakened . very bad relationship!
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cell is the smallest unit of life
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cell is the structural and functional unit of life
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Source:  OpenStax, Introductory statistics. OpenStax CNX. May 06, 2016 Download for free at http://legacy.cnx.org/content/col11562/1.18
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