There is sufficient evidence to support the claim that in a small city renowned for its music school, the average child takes at least 5 years of piano lessons

Step-by-step explanation:

From the question we are told that

The population mean is $\mu = 5 \ year$

The sample size is n = 20

The sample mean is $\= x = 4.6 \ years$

The standard deviation is $\sigma = 2.2 \ years$

The Null hypothesis is $H_o : \mu = 5$

The Alternative hypothesis is $H_a : \mu < 5$

So i will be making use of $\alpha = 0.05$ level of significance to test this claim

The critical value of $\alpha$ from the normal distribution table is $Z_\alpha = 1.645$

Generally the test statistics is mathematically evaluated as

$t = \frac{\= x - \mu}{ \frac{\sigma }{\sqrt{n} } }$

substituting values

$t = \frac{ 4.6 - 5}{ \frac{2.2}{\sqrt{20} } }$

$t = -0.8131$

Looking at the value of t and $Z_{\alpha }$ we see that $t < Z_{\alpha }$ so we fail to reject the null hypothesis

This implies that there is sufficient evidence to support the claim that in a small city renowned for its music school, the average child takes at least 5 years of piano lessons.

4 0

2 years ago

If angle symbol1 and angle symbol2 are complementary angles and if the measure of angle symbol1 is 5454degrees° more than the me

Misha Larkins [42]

Complementary angles by definition sum up to 90 degrees.
Let x = measure of angle2.
It's given that angle1 = x + 54

Since complementary, we know (x+54)+x = 90
2x+54=90
2x=36
x=18.
So angle2 measures 18 degrees and angle1 is 18 degrees+54 degrees, or 72 degrees. 72+18 = 90 as expected for complementary angles.

7 0

3 years ago