# Solved: Average adult Americans are about one inch taller, but nearly a whopping 25 pounds heavier than they were in 1960, according to a new report

#### ByDr. Raju Chaudhari

Feb 24, 2021

"Average adult Americans are about one inch taller, but nearly a whopping 25 pounds heavier than they were in 1960, according to a new report from the Centers for Disease Control and Prevention (CDC). The bad news, says CDC is that average BMI (body mass index, a weight-for-height forrnula used to measure obesity) has increased among adults from approximately 25 in 1960 to 28 in 2002." Boston is considered one of America's healthiest cities - is the weight gain since 1960 similar in Boston? A sample af n=25 adults suggested a mean increase of 17 pounds with a standard deviation of 8.6 pounds. Is Boston statistically significantly different in terms of weight gain since 1960? Run the appropriate test at a 5 % level of significance.

#### Solution

Given that the sample size $n = 25$, sample mean $\overline{x}= 17$ and sample standard deviation $s = 8.6$.

Hypothesis Testing Problem

The hypothesis testing problem is
$H_0 : \mu = 25$ against $H_1 : \mu \neq 25$ ($\text{two-tailed}$)

Test Statistic

The test statistic is
 \begin{aligned} t& =\frac{\overline{x} -\mu}{s/\sqrt{n}} \end{aligned}
which follows $t$ distribution with $n-1$ degrees of freedom.

Significance Level

The significance level is $\alpha = 0.05$.

Critical Value

As the alternative hypothesis is $\text{two-tailed}$, the critical value of $t$ for $24$ degrees of freedom $\text{are}$ $-2.064 and 2.064$.

The rejection region (i.e. critical region) is $\text{t < -2.064 or t > 2.064}$.

Computation

The test statistic under the null hypothesis is
 \begin{aligned} t&=\frac{ \overline{x} -\mu_0}{s/\sqrt{n}}\\ &= \frac{17-25}{8.6/ \sqrt{25 }}\\ &= -4.651 \end{aligned}

Decision

The test statistic is $t =-4.651$ which falls $\text{inside}$ the critical region, we $\text{reject}$ the null hypothesis.

$p$-value Approach

This is a $\text{two-tailed}$ test, so the p-value is the
area to the left of the test statistic ($t=-4.651$) is p-value = $10^{-4}$.

The p-value is $10^{-4}$ which is $\text{less than}$ the significance level of $\alpha = 0.05$, we $\text{reject}$ the null hypothesis.

We conclude that Boston is statistically significantly different in terms of weight gain since 1960.