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timurjin [86]
3 years ago
7

Taylor ran 57 minutes on Tuesday and 48 minutes on Saturday. Mark ran 62 minutes on Tuesday and 53 minutes on Saturday. Taylor s

aid they ran the same total time and mark said they didn’t. Who is right? Explain using comparative relational thinking.
Mathematics
2 answers:
marusya05 [52]3 years ago
8 0

Let's compare!

Taylor ran 57 + 48 mins total. Sum would be 105 mins.

Mark ran 62 + 53 mins total. Sum would be 115 mins.

Clearly, the two sums are different when we compare them using subtraction.

115 - 105 = 10. Mark ran 10 mins longer than Taylor.

vovikov84 [41]3 years ago
6 0

No they didn’t. So that makes mark right, because I’m an hour there is 60mins so Taylor’s time—57+48=105-60=45 that would be 1.45 (1hr45mins) Marks time— 62+53=115-60=55 so that would be 1.55(1hr55mins)

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Step-by-step explanation:

Data given

n_1 =6 represent the sample size for group 1

n_2 =5 represent the sample size for group 2

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\bar X_2 =38 represent the sample mean for the group 2

s_1=7 represent the sample standard deviation for group 1

s_2=8 represent the sample standard deviation for group 2

System of hypothesis

The system of hypothesis on this case are:

Null hypothesis: \mu_1 \leq \mu_2

Alternative hypothesis: \mu_1 > \mu_2

We are assuming that the population variances for each group are the same

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The statistic for this case is given by:

t=\frac{(\bar X_1 -\bar X_2)-(\mu_{1}-\mu_2)}{S_p\sqrt{\frac{1}{n_1}+\frac{1}{n_2}}}

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S^2_p =\frac{(n_1-1)S^2_1 +(n_2 -1)S^2_2}{n_1 +n_2 -2}

We can find the pooled variance:

S^2_p =\frac{(6-1)(7)^2 +(5 -1)(8)^2}{6 +5 -2}=55.67

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S_p=7.46

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The degrees of freedom are given by:

df=6+5-2=9

The p value is given by:

p_v =P(t_{9}>2.656) =0.0131

Since the p value is higher than the significance level given of 0.01 we don't have enough evidence to conclude that the true mean for group 1 is significantly higher thn the true mean for the group 2.

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