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kondor19780726 [428]

3 years ago

6

In a certain breed of cattle, the length of gestation has a mean of 284 days and a standard deviation is 5.5 days. One length of

gestation had a z-score of -1.70. Which of the following sentences best interprets the z-score?
A. The length of this gestation period was longer than the mean length of gestation periods by 1.7 days.
B. The length of this gestation period was shorter than the mean like the gestation periods by 1.7 days.
C. The length of this gestation period was longer than the mean length of gestation periods by 1.7 standard deviations.
D. The length of this gestation period was shorter than the mean length of gestation periods by 1.7 standard deviations.

2 answers:

Alenkinab [10]3 years ago

8 0

Answer:

The statement that best interprets the z-score is;

D. The length of the gestation period was shorter than the mean length of gestation period by 1.7 standard deviations

Step-by-step explanation:

The given parameters are;

The length of the gestation period of the cattle breed = 284 days

The standard deviation = 5.5 days

The z-score of one length of gestation period was z = -1.70

The formula for finding the z-score, z is presented as follows

$Z=\dfrac{x-\mu }{\sigma }$

Where;

x = The observed value for the gestation period

μ = The sample mean length for the gestation period = 284

σ = The gestation period standard deviation = 5.5

When Z = -1.70, we have;

$-1.70 =\dfrac{x-\mu }{\sigma }$

∴ x - μ = -1.7 × σ

∴ x is shorter than μ by 1.7×σ

Therefore, the length of the gestation period, x, was shorter than the mean length of gestation period, μ, by 1.7 standard deviations, σ.

nordsb [41]3 years ago

6 0

Answer:

D. The length of the gestation period was shorter than the mean length of gestation period by 1.7 standard deviations

Step-by-step explanation:

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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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