Answer:
Here, BRX and NJY are two triangles in which,
BR = 30 cm, RX = 40 cm, BX = 60 cm, NJ = 15 cm, JY = 20 cm and NY = 30 cm,
Also, m∠B = m∠N, m∠R = m∠J and m∠X = m∠Y,
By the property of congruence,
,
and
Thus, By AAA similarity postulate,
Hence, proved.
Answer:
The z-score (value of z) for an income of $1,100 is 1.
Step-by-step explanation:
We are given that the mean of a normally distributed group of weekly incomes of a large group of executives is $1,000 and the standard deviation is $100.
<em>Let X = group of weekly incomes of a large group of executives</em>
So, X ~ N()
The z-score probability distribution for a normal distribution is given by;
Z = ~ N(0,1)
where, = mean income = $1,000
= standard deviation = $100
The Z-score measures how many standard deviations the measure is away from the mean. After finding the Z-score, we look at the z-score table and find the p-value (area) associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X.
Now, we are given an income of $1,100 for which we have to find the z-score (value of z);
So, <em><u>z-score</u></em> is given by = =
= 1
<em>Hence, the z-score (value of z) for an income of $1,100 is 1.</em>
Answer:
a) Null hypothesis:
Alternative hypothesis:
b) the test statistics is : 2.15
c) The p-value is 0.0158
d) NO, there is evidence that the mean time of telephone survey is less than 15 and premium rate is not justified.
Step-by-step explanation:
The data in the Microsoft Excel are:
17;11;12;23;20;23;15;
16;23;22;18;23;25;14;
12;12;20;18;12;19;11;
11;20;21;11;18;14;13;
13;19; 16;10;22;18;23.
a) Formulate the null and alternative hypotheses for this application.
From the question, Fowle Marketing Research Inc. is taking base charge from a client on the given assumption that if the mean time of telephone survey is 15 minutes or less.
The null and alternative hypotheses are therefore as follows:
Null hypothesis:
The null hypothesis states that there is evidence that the mean time of telephone survey is less than 15 and premium rate is not justified.
Alternative hypothesis:
The alternative hypothesis states that there is evidence that the mean time of telephone survey exceeds 15 and premium rate is justified.
b) Compute the value of the test statistic.
Given that:
n = 35
The sample mean is;
Thus:
Thus; the test statistics is : 2.15
c) What is the p-value?
p-value = P(Z > 2.15)
p-value = 1 - P(Z ≤ 2.15)
From the standard normal table, the value of P(Z ≤ 2.15) is 0.9842
p-value = 1 - 0.9842
p-value = 0.0158
The p-value is 0.0158
d) At a = .01, what is your conclusion?
According to the rejection rule, if p-value is less than 0.01 then reject null hypothesis at ∝ = 0.01
Thus; p-value = 0.0158 > ∝ = 0.01
By the rejection rule, accept the null hypothesis.
Therefore, there is evidence that the mean time of telephone survey is less than 15 and premium rate is not justified.