A sample of size 60 from one population of weights had a sample average of 10.4 lb. and a sample standard deviation of 2.7 lb. An independent sample of size 100 from another population of weights had a sample average of 9.7 lb. with a sample standard deviation of 1.9 lb. Find a 95% confidence interval for the difference between the population means.

z=  0.278

Step-by-step explanation:

Given data

n1= 60 ; n2 = 100

mean 1= x1`= 10.4;     mean 2= x2`= 9.7

standard deviation  1= s1= 2.7 pounds ;  standard deviation 2= s2 = 1.9 lb

We formulate our null and alternate hypothesis as

H0 = x`1- x`2 = 0 and H1 = x`1- x`2 ≠ 0 ( two sided)

We set level of significance α= 0.05

the test statistic to be used under H0 is

z = x1`- x2`/ √ s₁²/n₁ + s₂²/n₂

the critical region is z > ± 1.96

Computations

z= 10.4- 9.7/ √(2.7)²/60+( 1.9)²/ 100

z= 10.4- 9.7/ √ 7.29/60 + 3.61/100

z= 0.7/√ 0.1215+ 0.0361

z=0.7 /√0.1576

z= 0.7 (0.396988)

z= 0.2778= 0.278

Since the calculated value of z does not fall in the critical region so we accept the null hypothesis H0 = x`1- x`2 = 0  at 5 % significance level. In other words we conclude that the difference between mean scores is insignificant or merely due to chance.

0.6, 3/5, and 6% are the 3 same portion.

but, when they're all changed to a decimal they would equal to 0.6.

hope this , good luck!

or

step-by-step explanation:

we need to solve for 'x' the following equation:  

we have that:

⇒

⇒ ⇒

factorizing, we get that:

  ⇒

then, the solutions are: or

nothing extraneous has been found!