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.
English translation:
If the quality control manager wants to estimate the average life of lamps with 20 hours, with a confidence level of 95% and an error of 10% is assumed. What is the sample size?
Answer:
x = 1 and y = 1
Step-by-step explanation:
7(1) minus 3(1) = 4
and 2(1) minus 1 = 1
just try and guess a number and try to solve the problem with that number, if it isn't the answer then repeat.
Answer:
2i: 169.71
2ii: 0.17L
3a: 4×10⁻⁵
3b: 110011
Step-by-step explanation:
2i. The surface of the top and bottom of the tin is two times (top and bottom) π·r² = 2·π·3² = 18π cm².
The circumference of the circle is 2·π·r = 6π cm².
The area of the material connecting top and bottom is a rectangle of the tin height times the circumference: 6·6π = 36π cm².
This gives a total of 18π + 36π = 54π cm².
With π approximated by 22/7 the total surface area is 54*22/7 ≈ 169.71.
Notice how the calculation is simple by waiting until the very last moment to substitute π.
2ii. The volume is the area π·r² of the circle times the height of the tin: 9π*6 = 54π cm³ ≈ 169.71 cm³.
Since 1L = 1000 cm³ the volume is 0.16971 litres, which should be rounded to 0.17 L.
3a: If we rewrite P as 36 x 10⁻⁴ and realize that 36/2.25 = 16, then the fraction can be written as
16 x 10⁻⁴⁻⁶ = 16 x 10⁻¹⁰.
The square root of that is taking it to the power of 1/2, so (16x10⁻¹⁰)^0.5 = 4x10⁻⁵ = 0.00004
3b: 1111 1111 is 255 in decimal. 101 is 5 in decimal. 255/5 is 51 in decimal. 51 in binary is 110011.
Tan=24/10 ≈ 2.4
using Pythagoras theorem
x²=(24)²+(10)²
x²=576+100
x²=57600
x=√57600 => 240
Therefore
sina=24/240 => 0.1
cota=1/tana => 1/(24/10) => 10/24
