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4 years ago

7

On the coldest night in Oymayakon, Russia, the temperature dropped from 0°C to −20°C in 5 hours. What is the average temperature

change per hour? Enter your answer in the box.

2 answers:

Andrej [43]4 years ago

4 0

it dropped from 0 to -20

-20 -0 = -20

divide by the time of 5 hours

-20/5

-4 degrees per hour

algol [13]4 years ago

4 0

average change per hour=(total change)/(time elapsed in hours)

total change is rom 0 to -20 or -20

hours=5

o average change is $\frac{-20^oC}{5hr}=-4 \frac{^oC}{hr}$

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A recent study focused on the number of times men and women who live alone buy take-out dinner in a month. Assume that the distr

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(a) Decision rule for 0.01 significance level is that we will reject our null hypothesis if the test statistics does not lie between t = -2.651 and t = 2.651.

(b) The value of t test statistics is 1.890.

(c) We conclude that there is no difference in the mean number of times men and women order take-out dinners in a month.

(d) P-value of the test statistics is 0.0662.

Step-by-step explanation:

We are given that a recent study focused on the number of times men and women who live alone buy take-out dinner in a month.

Also, following information is given below;

Statistic : Men Women

The sample mean : 24.51 22.69

Sample standard deviation : 4.48 3.86

Sample size : 35 40

<em>Let </em> $\mu_1$ <em> = mean number of times men order take-out dinners in a month.</em>

<em /> $\mu_2$ <em> = mean number of times women order take-out dinners in a month</em>

(a) So, Null Hypothesis, $H_0$ : $\mu_1-\mu_2$ = 0 {means that there is no difference in the mean number of times men and women order take-out dinners in a month}

Alternate Hypothesis, $H_A$ : $\mu_1-\mu_2\neq$ 0 {means that there is difference in the mean number of times men and women order take-out dinners in a month}

The test statistics that would be used here <u>Two-sample t test statistics</u> as we don't know about the population standard deviation;

T.S. = $\frac{(\bar X_1-\bar X_2)-(\mu_1-\mu_2)}{s_p \sqrt{\frac{1}{n_1}+\frac{1}{n_2} } }$ ~ $t__n_1_-_n_2_-_2$

where, $\bar X_1$ = sample mean for men = 24.51

$\bar X_2$ = sample mean for women = 22.69

$s_1$ = sample standard deviation for men = 4.48

$s_2$ = sample standard deviation for women = 3.86

$n_1$ = sample of men = 35

$n_2$ = sample of women = 40

Also, $s_p=\sqrt{\frac{(n_1-1)s_1^{2}+(n_2-1)s_2^{2} }{n_1+n_2-2} }$ = $\sqrt{\frac{(35-1)\times 4.48^{2}+(40-1)\times 3.86^{2} }{35+40-2} }$ = 4.16

So, <u>test statistics</u> = $\frac{(24.51-22.69)-(0)}{4.16 \sqrt{\frac{1}{35}+\frac{1}{40} } }$ ~ $t_7_3$

= 1.890

(b) The value of t test statistics is 1.890.

(c) Now, at 0.01 significance level the t table gives critical values of -2.651 and 2.651 at 73 degree of freedom for two-tailed test.

Since our test statistics lies within the range of critical values of t, so we have insufficient evidence to reject our null hypothesis as it will not fall in the rejection region due to which <u>we fail to reject our null hypothesis</u>.

Therefore, we conclude that there is no difference in the mean number of times men and women order take-out dinners in a month.

(d) Now, the P-value of the test statistics is given by;

P-value = P( $t_7_3$ > 1.89) = 0.0331

So, P-value for two tailed test is = 2 $\times$ 0.0331 = <u>0.0662</u>

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