Answer:
Since the computed value of t= 0.833 does not fall in the critical region we therefore do not reject H0 and may conclude that population mean is greater than 160. Or the sample comes from population with mean of 165.
Step-by-step explanation:
- State the null and alternative hypothesis as
H0: μ= 160 against the claim Ha :μ ≠160
Sample mean = x`= 165
Sample standard deviation= Sd= 12
2. The test statistic to use is
t= x`-μ/sd/√n
which if H0 is true , has t distribution with n-1 = 36-1= 35 degrees of freedom
3. The critical region is t< t (0.025(35)= 2.0306
t= x`-μ/sd/√n
4. t = (165-160)/[12/√(36)] = 5/[6] = 0.833
5. Since the computed value of t= 0.833 does not fall in the critical region we therefore do not reject H0 and may conclude that population mean is greater than 160. Or the sample comes from population with mean of 165.
Now
6. The p-value is 0 .410326 for t= 0.8333 with 35 degrees of freedom.
Answer:
Point
Explanation:
It makes the most sense
The best estimate of the population means is always the sample mean. We have the sample values, using them we can find the sample mean.
The sum of the sampled rates = 248
Sample Size = 10
Mean value of the sample = Sum of Sample Rate/Sample Size
So,
Mean of the sample = 248/10 = 24.8
Therefore, we can say that the best estimate for the population mean is 24.8
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Step-by-step explanation:
Answer:
look down below
Step-by-step explanation:
We know that AB = DC, and BC = AD, ∠B = ∠D because they are given
So, ΔABC ≅ ΔCDA because of SAS congruency
