Evaluate 5% of 123.22. Round to the nearest tenth.

A researcher was interested in comparing the resting pulse rates of people who exercise regularly and people who do not exercise

DiKsa [7]

Answer:

We Reject H₀ if t calculated > t tabulated

But in this case,

0.83 is not greater than 2.056

Therefore, we failed to reject H₀

There is no difference between the mean pulse rate of people who do not exercise and the mean pulse rate of people who do exercise.

Step-by-step explanation:

Refer to the attached data.

The Null and Alternate hypothesis is given by

Null hypotheses = H₀: μ₁ = μ₂

Alternate hypotheses = H₁: μ₁ ≠ μ₂

The test statistic is given by

$t = \frac{\bar{x}_1 - \bar{x}_2}{\sqrt{\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2} } }$

Where $\bar{x}_1$ is the sample mean of people who do not exercise regularly.

Where $\bar{x}_2$ is the sample mean of people who do exercise regularly.

Where $s_1$ is the sample standard deviation of people who do not exercise regularly.

Where $s_2$ is the sample standard deviation of people who do exercise regularly.

Where $n_1$ is the sample size of people who do not exercise regularly.

Where $n_2$ is the sample size of people who do exercise regularly.

$t = \frac{72.7 - 69.7}{\sqrt{\frac{10.9^2}{16} + \frac{8.2^2}{12} } }$

$t = 0.83$

The given level of significance is

1 - 0.95 = 0.05

The degree of freedom is

df = 16 + 12 - 2 = 26

From the t-table, df = 26 and significance level 0.05,

t = 2.056 (two-tailed)

Conclusion:

We Reject H₀ if t calculated > t tabulated

But in this case,

0.83 is not greater than 2.056

Therefore, We failed to reject H₀

There is no difference between the mean pulse rate of people who do not exercise and the mean pulse rate of people who do exercise.

4 0

3 years ago

Among a sample of 55 male soldiers the average shower time was found to be 2.95 minutes and the population standard deviation is

Oksanka [162]

Answer: -3.0593

Step-by-step explanation:

The test statistic for the difference between two population mean (when population standard deviations $\sigma_1\ \&\ \sigma_2$ are known ) is given by :-

$z=\dfrac{\overline{x}_1-\overline{x}_2}{\sqrt{\dfrac{\sigma_1^2}{n_1}+\dfrac{\sigma_2^2}{n_2}}}$

where ,

$n_1\ \&\ n_2$ = Sample sizes taken from two populations 1 and 2.

$\overline{x}_1-\overline{x}_2$ = Difference between two Sample mean.

$\sigma_1\ \&\ \sigma_2$ = Standard deviations of populations 1 and 2.

As per given , we have

$n_1=55\ \&\ n_2=58$

$\overline{x}_1=2.95\ \&\ \overline{x}_2=3.3$

$\sigma_1=0.65\ \&\ \sigma_2=0.56$

We assume that the data follows a normal distribution.

Then, the test statistic will be :-

$z=\dfrac{2.95-3.3}{\sqrt{\dfrac{(0.65)^2}{55}+\dfrac{(0.56)^2}{58}}}\\\\=\dfrac{-0.35}{\sqrt{0.0130887}}\\\\=\dfrac{-0.35}{0.1144059}=3.05928278174\approx-3.0593$

Hence, the test statistic= -3.0593

5 0

4 years ago

Solve for r. 1,2r-3=3(4-3,2r)

Helen [10]

Answer:

0.68865278

Step-by-step explanation:

7 0

3 years ago

I need some help with this

Korvikt [17]

The following can be deduced from the expressions

m ∠ D = 4 × m ∠ A
m ∠ B = 3 × m ∠ A-12

Please note that the sum of angles in a triangle is 180 degress. Therefore,

M∠ A + m ∠ B + m ∠ D = 180equation(3)

Substitute for equation 1 and 2 in equation 3 as shown below

m ∠ A + 3 × m ∠ A - 12^0 + 4 x m ∠ A =180^0
m ∠ A + 3 × m ∠ A + 4 x m ∠ A = 180^0+12^0
8m ∠ A = 192^0
m ∠ A= 192^0/8
m ∠ A=24^0

Substitute for m∠A= 24 in equation 1 and 2 as shown below

m ∠ D = 4 × m ∠ A
m ∠ D = 4 × 24^0
m ∠ D = 96^0
m ∠ B = 3 × mA - 12^0
m ∠ B = 3 × 24^0 - 12
m ∠ B = 72^0 - 12^0

,

∠= °

∠= °

∠= °

5 0

3 years ago

50 Points: Name the similar triangles & Prove the similarity with details

Leno4ka [110]

Let's consider the facts at hand:

By Vertical Angle Theorem ⇒ ∠BCE = ∠DCF
∠BEC = ∠DFC
Sides BE = DF

Based on the diagram, triangles BCE and triangles DCF are similar

⇒ based on the Angle-Angle theorem

⇒ since ∠BCE = ∠DCF and ∠BEC = ∠DFC

⇒ the two triangles are similar

Hope that helps!

Definitions of Theorem I used:

Vertical Angle Theorem: opposite angles of two intersecting lines must be equal
Angle-Angle Theorem: if two angles of both triangles are equal, then the given triangles must be similar

8 0

2 years ago